text:
\section{Proofs of the main results}
\label{sec:the-proof}

\subsection{Proof of the tube formula for fractal sprays, Theorem~\ref{thm:pointwise-tube-formula}}
\label{sec:the-proof-sprays}
\hfill \\
The proof of Theorem~\ref{thm:pointwise-tube-formula} is divided into several steps. We begin with a discussion intended to motivate and explain the approach.

  \begin{remark}[The philosophy behind the head and the tail]
  \label{rem:obs1}
  A technical part of the proof of Theorem~\ref{thm:pointwise-tube-formula} is inspired by the proof of the pointwise explicit formulas given in \cite[Chapter~5 and Section~ 8.1.1]{FGCD}. The main idea underlying the proof, however, is new and relies on the notions of ``head'' and ``tail''.

  In Section~\ref{sec:splitting}, we will split the tubular zeta function into a finite sum, which we call the head, and an infinite sum, which we call the tail (denoted by $\gzT[\head]$ and $\gzT[\tail]$, respectively); see \eqref{eqn:volume-zeta-split2}. Similarly, we will split the tube formula into corresponding finite and infinite sums ($V_{\mathrm{head}}$ and $V_{\mathrm{tail}}$, respectively); see \eqref{eqn:def:V-tail}. This decomposition will allow us to avoid repeating the same type of argument as appears in the proof of Proposition~\ref{prop:poles}, in several different instances. In particular, the decomposition into ``head'' and ``tail'' provides a technical device which allows us to use the Heaviside function as expressed in Lemma~\ref{lem:Heaviside-est}.

  The ``head'' and ``tail'' decomposition is justified by the observation that for every fixed $J\in\bN$, the complex dimensions of a fractal string $\sL = \{\scale_j\}_{j=1}^\iy$ do not depend on the first $J$ scaling ratios $\scale_1,\dots,\scale_J$. This is the idea underpinning Proposition~\ref{prop:poles}: the zeta function
  $\gz_{\sL_J}$ of the string $\sL_J := \sL\setminus\{\scale_1,\scale_2,\dots,
  \scale_J\}=\{\scale_j\}_{j=J+1}^\iy$ has the same poles as the zeta
  function \gzL of the full string \sL. Indeed, the function
  $f$ defined by $f(s)=\gr^s$, where $\gr>0$ is some positive real
  number, is entire, and so is any finite sum of such functions.
  Hence, if \gzL is meromorphic in some connected open neighborhood \gO of $W$, then
  since
  \[
  \gzL(s)=\sum_{j=1}^{J} \scale_j^s + \gz_{\sL_J}(s)
  \]
  for each $s\in\gO$, the truncated zeta function $\gz_{\sL_J}$ is
  meromorphic in \gO and a
  point $\pole \in \gO$ is a pole of \gzL if and only if it is a
  pole of $\gz_{\sL_J}$. Moreover, the residues of \gzL and
  $\gz_{\sL_J}$ at $\pole$ obviously coincide. The same applies if
  the first scaling ratios are changed instead of omitted.

  In addition to being technically useful, the ``head and tail'' decomposition helps one understand the conceptual difference between the contributions made to the tube formula by the integer and scaling dimensions, and the origin of the error term. Indeed, the bulk of the proof of Theorem~\ref{thm:pointwise-tube-formula} lies in showing that $V_{\mathrm{head}}$ is given by the residues of $\gzT[\head](\ge,s)$ at the integer dimensions, and $V_{\mathrm{tail}}$ is given by the residues of $\gzT[\tail](\ge,s)$ at the scaling dimensions; cf.~\eqref{eqn:head-term-assertion} and \eqref{eqn:tail-term-assertion}.
\end{remark}

  \subsubsection{Splitting the tubular zeta function.}
  \label{sec:splitting}

  For $k=0, 1, \dots, d$, we define the function $f_k:(0,\iy)\to \bR$ by
  \begin{equation}\label{eqn:def:f_k}
    f_k(\ge):= \crv_k(\gen,\ge)-\crv_k(\gen).
  \end{equation}
  This function measures the error of replacing the \kth coefficient function $\crv_k(\gen,\mydot)$ by the constant $\crv_k(\gen)$.
  Note that $f_k(\ge) = 0$ for $\ge \geq \genir$ and that in case $\crv_k(G,\mydot)$ is constant, $f_k \equiv 0$.

  By employing the functions $f_k$ defined above, the tubular zeta function \gzT can be rewritten in a more convenient way (see Definition~\ref{def:volume-zeta-fn}):
  \linenopax
  \begin{align}\label{eqn:volume-zeta-presplit}
    \gzT(\ge,s)

    &= \ge^{d-s} \sum_{j=1}^\iy \scale_j^s
    \sum_{k=0}^{d}\frac{\genir^{s-k} f_k(\scale_j^{-1}\ge)}{s-k} + \ge^{d-s}
    \sum_{j=1}^\iy \scale_j^s \left(\sum_{k=0}^{d} \frac{\genir^{s-k} \crv_k(\gen)}{s-k} -\frac{\genir^{s-d} \lambda_d(G)}{s-d}\right).
  \end{align}
  In the second term, it is now possible to separate the \emph{scaling zeta function} \gzL, since the sum over $k$ does not depend on $j$ any longer.
  Moreover, according to the fact that $\scale_j\to 0$ as $j\to\iy$, the first sum is taken only over finitely many integers $j$, namely, those indices for which $\scale_j \genir >\ge$. (In fact, it is the finiteness of the first sum which ensures that the expression on the right-hand side in \eqref{eqn:volume-zeta-presplit} converges absolutely exactly when the series defining $\gzT$ does. Hence, \eqref{eqn:volume-zeta-presplit} (as well as \eqref{eqn:volume-zeta-split2} below) holds for all $\ge>0$ and all $s\in\bC$ such that the second sum converges.)

  Set
  \linenopax
  \begin{align}\label{eqn:def-rho_j}
    \inradius_j := \gr(\gen^j) = \scale_j \genir,
  \end{align}
  and recall that $J=J(\ge)$ is the largest index such that
  $\inradius_J >\ge$, cf.~\eqref{eqn:J(eps)}.
  Obviously, $\inradius_j$ is the inradius of the set $G^j = \gY_j(\gen)$ and $G^j \ci (G^j)_{-\ge}$ iff $j > J$.
  Hence, for all $j > J$,
  \linenopax
  \begin{align*}
    \scale_j^{-1}\ge \geq \genir
    \q\q\text{and}\q\q
    f_k( \scale_j^{-1}\ge)=0 \q \text{ for } k=0,1,\dots, d.
  \end{align*}
  Interchanging the order of summation in the first term of \eqref{eqn:volume-zeta-presplit}, and making use of \eqref{eqn:leb-coeff-rel}, we conclude that the tubular zeta function is given by
  \linenopax
  \begin{align} \label{eqn:volume-zeta-split2}
    \gzT(\ge,s)
    &= \underbrace{\ge^{d-s} \sum_{k=0}^{d} \frac{\genir^{s-k}}{s-k} \sum_{j=1}^{J}\scale_j^s f_k(\scale_j^{-1}\ge)}_{=:\gzT[\head](\ge,s)} + \underbrace{\frac{\ge^{d-s}\gzL(s)}{d-s} \left(\sum_{k=0}^{d-1} \frac{\genir^{s-k}}{s-k} (d-k)\crv_k(\gen)\right)}_{=:\gzT[\tail](\ge,s)},
  \end{align}
  with $J=J(\ge)$, as in \eqref{eqn:J(eps)}. Note that the \nth[d] term in $\gzT[\tail](\ge,s)$ vanishes because of the presence of the factor $(d-k)$ inside the sum.

In combination with the splitting in \eqref{eqn:volume-zeta-split2}, Proposition~\ref{prop:poles} yields the following result.

\begin{theorem}
  \label{thm:Meromorphic-continuation-of-gzT}
  Assume that $W$ is a window for \gzL and that \gW is a connected open neighborhood of $W$ in which \gzL is meromorphic. Fix an arbitrary $\ge>0$. Then:
  \begin{ThmList}
    \item
    \emph{(Meromorphic continuation and poles of \gzT).} \;
     Both $\gzT(\ge,\mydot)$ and $\gzT[\tail](\ge,\mydot)$ are meromorphic in all of \gW. Furthermore, the set of visible poles of these functions is contained in $\sD_\tiling(W) = \DT \cap W$, as given in Definition~\ref{def:complex-dimn}.

    \item
    \emph{(Head and tail decomposition of $\gzT$).} \;
    For every $s \in \gW$ (in particular, for every $s \in W$), the meromorphic continuation of $\gzT(\ge,\mydot)$ to \gW is given by
    \linenopax
  \begin{align}\label{eqn:gzT-decomp}
      \gzT(\ge,s) = \gzT[\head](\ge,s) + \gzT[\tail](\ge,s)\,.
    \end{align}
    Here, $\gzT[\head](\ge,\mydot)$ is given by
    \linenopax
    \begin{align}\label{eqn:gzThead}
      \gzT[\head](\ge,s) = \ge^{d-s} \sum_{k=0}^d \frac{\genir^{s-k}}{s-k} \sum_{j=1}^{J(\ge)} \scale_j^s f_k(\scale_j^{-1}\ge),

    \end{align}
    which is meromorphic in all of \bC, with poles in $\{0,1,\dots,d\}$,
    where $f_k$ is as in \eqref{eqn:def:f_k} and $J(\ge) := \max\{j \geq 1 \suth \scale_j^{-1} \ge < \genir\} \vee 0$,
    \footnote{It may be useful to keep in mind that even though $J = J(\ge)$ is finite for every $\ge>0$, it tends monotonically to \iy as $\ge \to 0^+$, since $\scale_j$ decreases monotonically to $0$.}
    as in \eqref{eqn:J(eps)},
    while the meromorphic continuation of $\gzT[\tail](\ge,\mydot)$ to \gW is given by
    \linenopax
    \begin{align}\label{eqn:gzTtail}
      \gzT[\tail](\ge,s) = \frac{\ge^{d-s}}{d-s} \gzL(s) \sum_{k=0}^{d-1} \frac{\genir^{s-k}}{s-k} (d-k) \crv_k(\gen),

    \end{align}
    where $\gzL(s)$ denotes the meromorphic continuation of $\sum_{j=1}^\iy \scale_j^s$, as usual.

    \item
    \emph{(Residues of \gzT).} \;
    For each $\pole \in \sD_\sT(W) = (\DL \cup \{0,1,\dots,d\}) \cap W$, we have
    \linenopax
  \begin{align}\label{eqn:residue-zeta-split}
    \res{\gzT(\ge,s)}
    = \res{\gzT[\head](\ge,s)} + \res{\gzT[\tail](\ge,s)},
  \end{align}
   which implies
    \linenopax
    \begin{align}\label{eqn:res(gzt)=res(gzTtail)}
      \res{\gzT(\ge,s)} = \res{\gzT[\tail](\ge,s)}
    \end{align}
   for each $\pole \in \sD_\sL(W) \less \{0,1,\dots,d\}$, since $\gzT[\head](\ge,s)$ is holomorphic on $\bC \less \{0,1,\dots,d\}$. \\

    (i) In the case when $\pole \in \sD_\sL(W) \less \{0,1,\dots,d\}$ is a \emph{simple} pole of $\gzT(\ge,\mydot)$ (and hence also a simple pole of \gzL),
    \linenopax
    \begin{align}\label{eqn:value-of-res(gzT,w)}
      \res[\pole]{\gzT(\ge,s)}
      = \frac{\ge^{d-\pole}}{d-\pole} \res[\pole]{\gzL(s)} \sum_{k=0}^{d-1} \frac{\genir^{\pole-k}}{\pole-k} (d-k) \crv_k(\gen).
    \end{align}

    (ii) In the case when $\pole = k \in \{0,1,\dots,d\}$, we have
    \linenopax
    \begin{align}\label{eqn:value-of-res(gzThead,k)}
      \res[k]{\gzT[\head](\ge,s)}
      = \ge^{d-k} \sum _{j=1}^{J(\ge)} \scale_j^k f_k(\scale_j^{-1}\ge)\,.
    \end{align}
    Furthermore, if $\pole = k \in \{0,1,\dots,d-1\}$ is a \emph{simple} pole of $\gzT(\ge,\mydot)$, we also have
    \linenopax
    \begin{align}\label{eqn:value-of-res(gzTtail,k)}
      \res[k]{\gzT[\tail](\ge,s)}
      = \ge^{d-k} \gzL(k) \crv_k(\gen).
    \end{align}
    If $\pole = d \in W$ and $\abscissa < d$ (as assumed in Theorem~\ref{thm:pointwise-tube-formula}), then $d$ is not a pole of \gzL. Hence, formulas \eqref{eqn:gzTtail} and \eqref{eqn:leb-coeff-rel} imply that
    \linenopax
    \begin{align}\label{eqn:res(gzT,d)}
      \res[d]{\gzT[\tail](\ge,s)} = \gzL(d) \left(\crv_d(\gen) - \gl_d(\gen)\right)\,.
    \end{align}
  \end{ThmList}
\end{theorem}

We leave it to the interested reader to perform the necessary (and elementary) computations needed to deal with the case when \pole is a multiple pole of \gzT; see \cite[Section~6.1.1]{FGCD}.

\begin{remark}\label{rem:integer-residue-as-Laurent-constant}
  If we assume that $\pole \in \{0,1,\dots,d-1\} \cap \sD_\sL(W)$ in part 3 of Theorem~\ref{thm:Meromorphic-continuation-of-gzT}, then one can express $\res{\gzT[\tail](\ge,s)}$ in terms of the constant term of the Laurent expansion of $\gzL(s)$ at $s=k$ (even in the case when $k$ is a multiple pole of \gzL), much as was done when $d=1$ in \cite[Corollary~8.10 and Remark~8.11]{FGCD}.
\end{remark}

  \subsubsection{Splitting the parallel volume}

  For fixed $\ge > 0$, we split the inner parallel volume $V(\tiling,\ge)$ in a similar way as the tubular zeta function above.
  Taking into account first \eqref{eqn:V(Gj,ge)} and then \eqref{eqn:def:f_k}, we can use \eqref{eqn:def-rho_j} to write
  \linenopax
  \begin{align} \label{eqn:split-T-volume}
  V(\tiling,\ge)
    &= \sum_{j: \inradius_j >  \ge} V(\gen^j,\ge) + \sum_{j: \inradius_j \leq  \ge} \gl_d(\gen^j) \notag \\
    &= \sum_{j: \inradius_j > \ge} \sum_{k=0}^{d} \ge^{d-k} \scale_j^k \crv_k(\gen,\scale_j^{-1}\ge)
     + \sum_{j: \inradius_j \leq  \ge} \gl_d(\gen^j)\\
    &= \sum_{j: \inradius_j >  \ge} \sum_{k=0}^{d} \ge^{d-k} \scale_j^k
      f_k(\scale_j^{-1}\ge)
    + \sum_{j: \inradius_j >  \ge} \sum_{k=0}^{d} \ge^{d-k}
      \scale_j^k \crv_k(\gen) + \sum_{j: \inradius_j \leq  \ge}
      \gl_d(\gen^j)
      .
      \notag
  \end{align}

  Recall that the sum over $j$ in the first two terms is finite for each fixed $\ge > 0$ and that the number of terms is given by $J=J(\ge)$.
  Therefore, in both terms, the sums can be interchanged.
  In the third term, the homogeneity of the $d$-dimensional Lebesgue measure $\gl_d$ implies $\gl_d(\gen^j)=\gl_d(\scale_j \gen)=\scale_j^d \gl_d(\gen)$ for each
  $j$.
  Thus \eqref{eqn:leb-coeff-rel} yields
  \linenopax
  \begin{align}\label{eqn:def:V-tail}
    V(\tiling,\ge)
    &= \sum_{k=0}^{d} \ge^{d-k} \sum_{j=1}^{J} \scale_j^k f_k(\scale_j^{-1}\ge) + \sum_{k=0}^{d} \crv_k(\gen) \negsp[10] \sum_{j: \inradius_j >  \ge} \ge^{d-k} \scale_j^k
      \; + \; \gl_d(\gen) \negsp[10] \sum_{j: \inradius_j \leq  \ge} \scale_j^d \notag \\
        &= \underbrace{\sum_{k=0}^{d} \ge^{d-k} \sum_{j=1}^{J} \scale_j^k f_k(\scale_j^{-1}\ge)}_{=:V_{\mathrm{head}}(\tiling,\ge)} + \underbrace{\sum_{k=0}^{d}\crv_k(\gen) \left(\sum_{j: \inradius_j >  \ge} \ge^{d-k} \scale_j^k
      + \sum_{j: \inradius_j \leq  \ge} \genir^{d-k} \scale_j^d\right)}_{=:V_{\mathrm{tail}}(\tiling,\ge)}.
  \end{align}

  \subsubsection{Outline of the remainder of the proof of the tube formula}

  In light of \eqref{eqn:def:V-tail}, the tube formula \eqref{eqn:pointwise-tube-formula} of Theorem~\ref{thm:pointwise-tube-formula} will follow upon verification of the following two assertions:
  \linenopax
  \begin{align}
    V_{\mathrm{head}}(\tiling,\ge)
    &= \sum_{\pole\in\sD_{\tiling}(W)} \negsp[8] \res[s=\pole]{\gzT[\head](\ge,s)}+\R_{\he}(\ge)\,, \q\text{and}
    \label{eqn:head-term-assertion} \\
    V_{\mathrm{tail}}(\tiling,\ge)
    &= \negsp[5] \sum_{\pole\in\sD_{\tiling}(W)} \negsp[8] \res[s=\pole]{\gzT[\tail](\ge,s)} +  \gl_d(G)\gzL(d) + \R_{\ta}(\ge),
    \label{eqn:tail-term-assertion}
  \end{align}
  where the error terms $\R_{\he}$ and $\R_{\ta}$ are given by
  \linenopax
  \begin{align}
    \R_{\he}(\ge) \label{eqn:pointwise-error-head}
    &= \sum_{k=0}^{[S(0)]} \res[s=k]{\gzT[\head](\ge,s)}\,, \q\text{and}\\
    \R_{\ta}(\ge) \label{eqn:pointwise-error-head-tail}
    &= \frac{1}{2\gp\ii} \int_S \gzT[\tail](\ge,s)\,ds\,,

  \end{align}
  respectively. (Here $[x]$ denotes the integer part of $x$.)
  Indeed, the assumption $S(0)<0$ implies immediately $\R_{\he}\equiv 0$ and, by~\eqref{eqn:pointwise-error},  we have $\R(\ge)= \R_{\ta}(\ge)$. Therefore, if the formulas (\ref{eqn:head-term-assertion}) and
  (\ref{eqn:tail-term-assertion}) hold, then  since $\gzT(\ge,s) = \gzT[\head](\ge,s) + \gzT[\tail](\ge,s)$  by \eqref{eqn:volume-zeta-split2}, it follows from \eqref{eqn:def:V-tail} that
  \linenopax
  \begin{align*}
    V(\tiling,\ge)
    &= V_{\mathrm{head}}(\tiling,\ge) + V_{\mathrm{tail}}(\tiling,\ge)  \\
    &=\negsp[15]\sum_{\pole\in\sD_{\tiling}(W)} \negsp[10] \res[s=\pole]{\gzT[\head](\ge,s)}+ \R_{\he}(\ge)\\
    &\hstr[10]+ \negsp[15]\sum_{\pole\in\sD_{\tiling}(W)} \negsp[10]\res[s=\pole]{\gzT[\tail](\ge,s)} +  \gl_d(G)\gzL(d)+ \R_{\ta}(\ge)\\
    &= \negsp[10]\sum_{\pole\in\sD_{\tiling}(W)} \negsp[10]\res[s=\pole]{\gzT[\head](\ge,s)+\gzT[\tail](\ge,s)} +  \gl_d(G)\gzL(d) + \R(\ge)\\
    &= \negsp[10]\sum_{\pole\in\sD_{\tiling}(W)} \negsp[10]\res[s=\pole]{\gzT(\ge,s)} +  \gl_d(G)\gzL(d) + \R(\ge)\,,
  \end{align*}
  and we obtain the tube formula \eqref{eqn:pointwise-tube-formula}, with $\R(\ge)$ given as in
  formula \eqref{eqn:pointwise-error}:
  \linenopax
  \begin{align}\label{eqn:Rhead-expr}
    \R(\ge)
    = \R_{\ta}(\ge)
    = \frac1{2\gp\ii} \int_S \gzT[\tail](\ge,s) \,ds
  \end{align}
  Observe that in light of \eqref{eqn:volume-zeta-split2}, the integrand $\gzT[\tail](\ge,s)$ coincides with that of \eqref{eqn:pointwise-error}.

  We note that in order to complete the proof of Theorem~\ref{thm:pointwise-tube-formula}, we will still have to establish that the error term $\R_{\ta}(\ge)$ (which coincides with the full error term $\R(\ge)$, since $\R_{\he}(\ge)\equiv 0$, as noted in \eqref{eqn:Rhead-expr}) satisfies the asymptotic estimate $\R_{\ta}(\ge)=O(\ge^{d-\sup S})$ as $\ge\to 0^+$ (in the languid case) and that $\R_{\ta}(\ge)=0$ for all $0<\ge<\min\{g,A^{-1}g\}$ (in the strongly languid case). This will be accomplished, respectively, in Section~\ref{subsec:proof-error-term} and Section~ \ref{subsec:proof-strongly-languid}.

  We will establish independently (in Section~\ref{subsec:proof-head} and Section~ \ref{subsec:proof-tail}, respectively) that $V_{\mathrm{head}}(\tiling,\ge)$ can be expressed as a sum of residues of $\gzT[\head](\ge,s)$ (as given by \eqref{eqn:head-term-assertion}) and $V_{\mathrm{tail}}(\tiling,\ge)$ as a sum of residues of $\gzT[\tail](\ge,s)$ (as given by \eqref{eqn:tail-term-assertion}).
  While $\gzT[\head](\ge,\mydot)$ is meromorphic in $\bC$ and has poles only at the integer values $0,1,\dots, d$, the function $\gzT[\tail](\ge,s)$ recovers the full set of the complex dimensions of \sL (compare Theorem~\ref{thm:Meromorphic-continuation-of-gzT}).
  This fits perfectly with the observation (compare Remark~\ref{rem:obs1}) that the first lengths of a fractal string (or, as here, the first scaled copies of \gen in a fractal spray) do not affect its complex dimensions.
  The scaling complex dimensions are hidden in the ``tail''.
  The derivation of the first part is elementary and exact in that the error term just collects the residues at the integer dimensions not contained in $W$.
  The derivation of the second part is more involved.
  It uses techniques similar to those used in the proof of \cite[Theorem~8.7]{FGCD}, the pointwise tube formula for (1-dimensional) fractal strings.
  Here, it is necessary for \sL to be languid (for the first part of Theorem~\ref{thm:pointwise-tube-formula}) or strongly languid (for the second part).

  \subsubsection{Proof of \eqref{eqn:head-term-assertion}} \label{subsec:proof-head}

  The residue of $\gzT[\head](\ge,s)$ at $s=k \in \{0,1,\dots,d\}$ is given by formula \eqref{eqn:value-of-res(gzThead,k)} in Theorem~\ref{thm:Meromorphic-continuation-of-gzT}. Observing that the $k$-th term of $V_{\mathrm{head}}(\tiling,\ge)$ in \eqref{eqn:def:V-tail} has exactly the same expression, we conclude that

  \linenopax
  \begin{align*}
    V_{\mathrm{head}}(\tiling,\ge)
    &= \sum_{k=0}^d\res[s=k]{\gzT[\head](\ge,s)}.
  \end{align*}
For a fixed screen $S$ (and a corresponding window $W$) such that $S(0)\not\in\{0,\ldots,d\}$, we can split this sum into two parts, according to whether $k$ is contained in the interior of $W$ or in the complement $W^c$.
Since $\gzT[\head](\ge,\mydot)$ has no poles outside the set $\{0,\ldots,d\}$, we can safely extend the first sum to include the residues at all complex dimensions visible in $W$. Thus
  \linenopax
  \begin{align*}
    V_{\mathrm{head}}(\tiling,\ge)
    &=\negsp[15]\sum_{\pole\in\sD_{\tiling}(W)} \negsp[10] \res[s=\pole]{\gzT[\head](\ge,s)}+ \negsp[15]\sum_{k\in\{0,\ldots,d\}\cap W^c} \negsp[10] \res[s=k]{\gzT[\head](\ge,s)}\,,
  \end{align*}
 where the second sum is $\R_{\he}(\ge)$ (as defined in \eqref{eqn:pointwise-error-head}). This completes the proof of \eqref{eqn:head-term-assertion}.
We note that in the first part of Theorem~\ref{thm:pointwise-tube-formula}, the assumption $S(0)<0$ for the screen ensures that $\R_{\he}\equiv 0$. In Corollary~\ref{thm:fractal-spray-tube-formula-monophase}, where this assumption is dropped, we also have $\R_{\he}\equiv 0$, but for a different reason; indeed, in the monophase case, $\gzT[\head](\ge,s)$ itself vanishes identically. In the general case (which is not treated in this paper), when the integer dimensions may not all be visible (i.e., $S(0)>0$) and when $G$ is not necessarily monophase, the error term $\R_{\he}$ will have to be taken into account; see also Section~\ref{rem:0-in-W-redux}.

  \subsubsection{The Heaviside function}

  Before continuing on to the proof of \eqref{eqn:tail-term-assertion}, we need to make some remarks on a certain useful form of the \emph{Heaviside function} $H:\bR\to \{0,1\}$, which is defined (as in \cite[(5.10)]{FGCD}) by
  \linenopax
  \begin{align}\label{eqn:H(x)}
    H(x) =
    \begin{cases}
      0, &x<0 \\
      \tfrac12,& x=0, \\
      1, &x>0.
    \end{cases}
  \end{align}

  Following the proof of \cite[Theorem~8.7]{FGCD},
  we will exploit the following integral representation of the Heaviside function, which comes from number theory \cite[p.105]{Davenport}, and was refined in \cite[Lemma~5.1]{FGCD}.

  \begin{lemma}\label{lem:Heaviside-est}
  For $x,y,c>0$, the Heaviside function is given by
  \linenopax
  \begin{equation}\label{eqn:cor:Heavi-est}
    H(x-y) = \frac{1}{2\gp\ii} \int_{c-\ii \iy}^{c+\ii \iy} x^{s} y^{-s} \frac{ds}{s}.
    \end{equation}
  \begin{proof}
	Taking $k=1$ in \cite[Lemma~5.1]{FGCD} for a screen with
$T=T_+=-T_- >0$, the Heaviside function is approximated by
	\linenopax
	\begin{equation}\label{eqn:Heaviside-est}
	  H(x-y) = \frac{1}{2\gp\ii} \int_{c-\ii T}^{c+\ii T} x^{s}
	  y^{-s}\frac{ds}{s} + E,
	\end{equation}
	where the absolute value of the error term $E$ is bounded by
	\linenopax
	\begin{equation}\label{eqn:Heaviside-error-est}
	  \begin{cases}
	    \left(xy^{-1}\right)^c \frac{1}{T} \min\{T,|\log(xy^{-1})|^{-1}\},
	      &\mbox{ if } x \neq y, \\
	    \frac{c}{T},
	      &\mbox{ if } x=y.
	    \end{cases}
	\end{equation}
	It is now easily seen, that, for arbitrary fixed values
	$x,y,c>0$, the error term $E = E(T)$ vanishes as $T \to \iy$.
	Since this is true for all $x,y,c>0$, the result follows.
  \end{proof}
  \end{lemma}

  \subsubsection{Proof of \eqref{eqn:tail-term-assertion}}
  \label{subsec:proof-tail}
  The proof of this part follows roughly the lines of the proof of \cite[Theorem~8.7]{FGCD}.
  One can rewrite the expression of $V_{\mathrm{tail}}(\tiling,\ge)$ in \eqref{eqn:def:V-tail} as
  \linenopax
  \begin{equation} \label{eqn:tail-term2}
    V_{\mathrm{tail}}(\tiling,\ge)
    = \sum_{k=0}^{d-1} \genir^{-k} \crv_k(\gen) \left(\sum_{j: \gr_j \geq \ge} \ge^{d-k} \gr_j^k
    + \sum_{j: \gr_j < \ge}  \gr_j^d\right) + \crv_d(G) \gzL(d).
  \end{equation}
  For $k=0, 1, \dots, d-1$, denote the expression within the parentheses of \eqref{eqn:tail-term2} by $v_k(\ge)$.

  Using the Heaviside function as defined in \eqref{eqn:H(x)}, we write
  \linenopax
  \begin{align} \label{eqn:v_k-split}
    v_k(\ge)
    &= \sum_{j: \gr_j  \geq \ge}  \ge^{d-k} \gr_j^k  +  \sum_{j: \gr_j  < \ge} \gr_j^d
    = \sum_{j=1}^\iy \left[\ge^{d-k} \gr_j^k H(\gr_j -\ge) +  \gr_j^d H(\ge-\gr_j )\right].
  \end{align}

  Note that, in case $\ge = \gr_j $ for some $j$, the corresponding \jth term in the sum on the right-hand side of \eqref{eqn:v_k-split} is
  \linenopax
  \begin{align*}
    \ge^{d-k} \gr_j^k H(\gr_j -\ge) +  \gr_j^d H(\ge-\gr_j )
    = \tfrac{1}{2} \gr_j^{d-k} \gr_j^k + \tfrac{1}{2} \gr_j^d
    =  \gr_j^d,
  \end{align*}
  which equals the value given by the left-hand side of \eqref{eqn:v_k-split}.

  Now, fix some constant $c$ such that $d-1<c<d$ and $\abscissa<c$.
  (This is possible, since the abscissa of convergence $\abscissa$
  of \gzL was assumed to be strictly less than $d$.) Then
  $c-k$ and $d-c$ are positive numbers and so, by
  Lemma~\ref{lem:Heaviside-est}, the $j$-th term in the above sum
  is given by
  \linenopax
  \[
  \int_{c-k-\ii \iy}^{c-k+\ii \iy} \ge^{d-k-t} \gr_j^{k+t}
  \frac{dt}{2\gp\ii t} +\int_{d-c-\ii \iy}^{d-c+\ii \iy}  \ge^t
  \gr_j^{d-t} \frac{dt}{2\gp\ii t}.
  \]
  Substituting $s=t+k$ in the first integral and $s=d-t$ in the
  second one, we get
  \linenopax
  \[
  \int_{c-\ii \iy}^{c+\ii \iy} \ge^{d-s} \gr_j^{s}
  \frac{ds}{2\gp\ii (s-k)} +\int_{c-\ii \iy}^{c+\ii \iy}  \ge^{d-s}
  \gr_j^{s} \frac{ds}{2\gp\ii (d-s)}.
  \]
  Combining these integrals, one sees that \eqref{eqn:v_k-split} can be rewritten as follows:

  \linenopax
  \begin{align} \label{eqn:v-k-2}
    v_k(\ge)

    &= \frac{d-k}{2\gp}  \sum_{j=1}^\iy\int_{-\iy}^{\iy}
       \frac{\ge^{d-c-\ii t} \gr_j^{c+\ii t}}{(c-k+\ii t)(d-c-\ii t)}dt.
  \end{align}

  This last integral converges absolutely, which is seen
  as follows. Since $c$ was chosen strictly between $d-1$ and
  $d$, the numbers $d-c$ and $c-k$ are positive and so we have
  for all real numbers $t$,
  \linenopax
  \[
  |(c-k+\ii t)(d-c-\ii t)|= |(c-k)(d-c)+ t^2+ \ii (d-k)t|\geq
  (c-k)(d-c)+ t^2.
  \]
  Hence
  \linenopax
  \begin{align*}
    \left|\frac{\ge^{d-c-\ii t} \gr_j^{c+\ii t}}{(c-k+\ii t)(d-c-\ii   t)}\right|
    &= \frac{\ge^{d-c} \gr_j^{c}|\ge^{-\ii t}| |\gr_j^{\ii   t}|}{|(c-k+\ii t)(d-c-\ii t)|}
    \le  \frac{\ge^{d-c} \gr_j^{c}}{(c-k)(d-c)+t^2},
  \end{align*}
  which implies
  \linenopax
  \begin{align*}
    \frac{1}{2\gp } \sum_{j=1}^\iy\int_{-\iy}^{\iy}
      \left|\frac{\ge^{d-c-\ii t} \gr_j^{c+\ii t}}{(c-k+\ii t)(d-c-\ii t)}\right|dt
    &\leq  \frac{1}{2\gp} \sum_{j=1}^\iy \gr_j^{c}\ge^{d-c} \int_{-\iy}^{\iy} \frac{dt}{(c-k)(d-c)+t^2} \\
    &=  \frac{1}{\gp } \gzL(c) \ge^{d-c} \int_{0}^{\iy}
    \frac{dt}{(c-k)(d-c)+t^2}<\iy.
  \end{align*}
  Note that the last expression is finite because $c$ was chosen such that
  $c>\abscissa$, where \abscissa is the abscissa of convergence of \gzL.

  It follows that the integrand in (\ref{eqn:v-k-2}) is absolutely integrable
  and it is safe to interchange the order of summation and
  integration. Hence
  \linenopax
  \begin{align*}
   v_k(\ge)
   &= \frac{1}{2\gp \ii} \int_{c-\ii \iy}^{c+\ii \iy}
      \ge^{d-s} \sum_{j=1}^\iy  \gr_j^{s} \frac{(d-k) }{(s-k)(d-s)}ds,
  \end{align*}
  where the sum $\sum_{j=1}^\iy \gr_j^{s}$ can be replaced by
  $\genir^s \gzL(s)$.

  \begin{comment}
  Separating now the two fractions we merged previously, we end
  up with
  \linenopax
  \begin{align*}
    v_k(\ge)
    &= \frac{1}{2\gp \ii} \int_{c-\ii \iy}^{c+\ii \iy}
       \ge^{d-s} \gzL(s) \left(\frac{1}{s-k} + \frac{1}{d-s}\right) ds\\
    &= \frac{1}{2\gp \ii} \int_{c-\ii \iy}^{c+\ii \iy}
       \ge^{d-s} \gzL(s) \frac{ds}{s-k}
       + \frac{1}{2\gp \ii} \int_{c-\ii \iy}^{c+\ii \iy}
         \ge^{d-s} \gzL(s) \frac{ds}{d-s}.
  \end{align*}
  \end{comment}

  Inserting the derived expressions for $v_k(\ge)$ into
  (\ref{eqn:tail-term2}), we obtain
  \linenopax
  \begin{align*}
    V_{\mathrm{tail}}(\tiling,\ge)

    &= \frac{1}{2\gp \ii} \sum_{k=0}^{d-1} \crv_k(\gen) \int_{c-\ii \iy}^{c+\ii \iy}
       \ge^{d-s} \gzL(s) \frac{\genir^{s-k} (d-k)}{(s-k)(d-s)}ds +  \crv_d(G)\gzL(d)\\
    &= \frac{1}{2\gp \ii} \int_{c-\ii \iy}^{c+\ii \iy}
       \frac{\ge^{d-s} \gzL(s)}{d-s}\sum_{k=0}^{d-1} \frac{\genir^{s-k}}{s-k} (d-k)\crv_k(\gen) ds +  \crv_d(G)\gzL(d) \\
    &= \frac{1}{2\gp \ii} \int_{c-\ii \iy}^{c+\ii \iy}
       \gzT[\tail](\ge,s) \, ds +  \crv_d(G)\gzL(d),
  \end{align*}
  by \eqref{eqn:volume-zeta-split2}.

  Since $d$ is not a pole of $\gzL$, it is a simple pole of $\gzT[\tail](\ge,\mydot)$

  and \eqref{eqn:res(gzT,d)} yields
  \linenopax
  \begin{align}\label{eqn:tail-residue-at-d}
    \res[s=d]{\gzT[\tail](\ge,s)}=\gzL(d)(\gk_d(G)-\gl_d(G)).
  \end{align}
  Therefore, we obtain
  \linenopax
  \begin{align*}
    V_{\mathrm{tail}}(\tiling,\ge)
    &= \frac{1}{2\gp \ii} \int_{c-\ii \iy}^{c+\ii \iy} \gzT[\tail](\ge,s) ds +  \res[s=d]{\gzT[\tail](\ge,s)}+\gzL(d)\gl_d(G).
  \end{align*}

  Now the machinery of the Residue Theorem can be applied.
  When pushing the line of integration towards the screen $S$, we
  collect on the way the residues of the poles of
  $\gzT[\tail](\ge,\mydot)$ that lie between the line $\Re s=c$
  and $S$ (see the proof of \cite[Theorem~8.7]{FGCD}). The
  definition of the screen $S$ and the window $W$ imply that
  \gzL is meromorphic in $W$ and, since $\abscissa < c$,
  \gzL has no poles to the right of the vertical line $\Re
  s=c$. Therefore, by Theorem~\ref{thm:Meromorphic-continuation-of-gzT}, any pole
  of $\gzT[\tail](\ge,\mydot)$ in the region between $\Re s=c$
  and $S$ is either contained in $\{0, 1, \dots, d-1\}$ or a pole
  of \gzL in $W$, i.e., an element of $\sD_{\tiling}(W)\setminus\{d\}$.

  Recall that $\gzT[\tail](\ge,\mydot)$ has another pole at
  $d$ but, since $d$ is not passed when pushing the line of
  integration towards the screen, it does not occur again.

  At this point, the languidness of \gzL comes into play.
  Using the sequence $\{T_n\}_{n\in\bZ}$ of Definition~\ref{def:languid}, we write
  $V_{\mathrm{tail}}(\tiling,\ge)$ as a limit of truncated
  integrals:
  \linenopax
  \begin{equation} \label{eqn:V-tail-limit}
    V_{\mathrm{tail}}(\tiling,\ge)
    = \lim_{n\to\iy} \frac{1}{2\gp \ii} \int_{c+\ii T_{-n}}^{c+\ii T_n}
      \gzT[\tail](\ge,s) ds +  \res[s=d]{\gzT[\tail](\ge,s)}+ \gzL(d)\gl_d(G).
  \end{equation}
  If we now replace the vertical line segment $\sC_{|n}:=[c+\ii T_{-n},c+\ii T_n]$
  of integration by the curve given by the union of the two
  horizontal line segments and the truncated screen $S_{|n}$, that is,
  \linenopax
  \begin{align*}
    U_{|n} &:= [c+\ii T_n, S(T_n)+\ii T_n], \\
    \genstr_{|n} &:= [c+\ii T_{-n}, S(T_{-n})+\ii T_{-n}], \q\text{and} \\
    S_{|n} &:= \{S(t)+\ii t: t\in [T_{-n},T_n]\}
  \end{align*}
  with proper orientations, the Residue Theorem implies that
  the \nth integral in \eqref{eqn:V-tail-limit} is equal to
  \linenopax
  \begin{equation*}
    \sum_{\pole\in\sD(W_{|n})} \res[s=\pole]{\gzT[\tail](\ge,s)}+ \sR_n(\ge) +
    U_n^\integral(\ge) + L^\integral_n(\ge),
  \end{equation*}
  where $\sD(W_{|n}):=\sD_{\tiling}(W)\cap W_{|n}$ is the set of
  (possible) poles of $\gzT[\tail](\ge,\mydot)$ that lie inside
  the region $W_{|n}$ bounded by the curves $U_{|n}, S_{|n}, \genstr_{|n}$ and
  $\sC_{|n}$. Hence, $W_{|n}$ is the ``truncated window'' associated to the truncated screen $S_{|n}$. The term $\sR_n(\ge)$ is given by the integral
  \linenopax
  \begin{equation} \label{error1}
    \sR_n(\ge)= \frac{1}{2\gp \ii} \int_{S_{|n}} \gzT[\tail](\ge,s) \, ds
  \end{equation}
  and $U_n^\integral(\ge)$ and $\genstr_n^\integral(\ge)$ are the corresponding integrals
  over the segments $U_{|n}$ and $\genstr_{|n}$, respectively (traversed clockwise around $W_{|n}$).

  More precisely,
  \smash{$U_n^\integral(\ge)$} is given by
  \linenopax
  \begin{align*}
    U_n^\integral(\ge)
    &= \frac{1}{2\gp \ii} \int_{S(T_n)+\ii T_n}^{c+\ii T_n} \gzT[\tail](\ge,s) \,ds\\
    &= \frac{1}{2\gp \ii} \int_{S(T_n)}^{c} \ge^{d-t-\ii T_n} \gzL(t+\ii T_n)
       \sum_{k=0}^{d-1} \frac{\genir^{t+\ii T_n-k} \crv_k(\gen)(d-k) }{(t-k+\ii T_n)(d-t-\ii T_n)} dt
  \end{align*}
  and is absolutely bounded as follows:
  \linenopax
  \begin{align*}
    |U_n^\integral(\ge)|

    &\leq \frac{1}{2\gp} \int_{S(T_n)}^{c} \ge^{d-t} |\gzL(t+\ii T_n)|
      \sum_{k=0}^{d-1} \frac{|\genir^{t-k} \crv_k(\gen)|(d-k) }{|t-k+\ii T_n||d-t-\ii T_n|} dt.
  \end{align*}
  According to the languidness condition \textbf{\ref{eqn:L1}} of Definition~\ref{def:languid} and the hypotheses of the first part of Theorem~\ref{thm:pointwise-tube-formula},

  there exist constants $C>0$ and
  $\languidOrder<1$, such that $|\gzL(t+\ii T_n)|\le C (T_n +1)^{\languidOrder}$.
  Moreover, $|t-k+\ii T_n|\geq T_n$, for all $k=0,\dots,d-1$ and,
  similarly, $|d-t-\ii T_n|\geq T_n$. Hence we get
  \linenopax
  \begin{equation} \label{u-n1}
    |U_n^\integral(\ge)| \le \frac{1}{2\gp} C (T_n +1)^{\languidOrder} \sum_{k=0}^{d-1}
  \frac{|\crv_k(\gen)|(d-k) }{T_n^2} \int_{S(T_n)}^{c} \genir^{t-k} \ge^{d-t} dt.

  \end{equation}
  Since $S(T_n) \geq \inf S$, the integral in this expression is
  bounded by a constant independent of $n$. Thus, there is a
  constant $C_1>0$, independent of $n$, such that
  \linenopax
  \begin{equation}\label{u-n2}
    |U_n^\integral(\ge)| \le C_1 \frac{(T_n +1)^{\languidOrder}}{T_n^2}.
  \end{equation}
  With similar arguments, one can show that the integral
  $\genstr_n^\integral(\ge)$ is absolutely bounded by $C_2 |T_{-n}|^{-2}(|T_{-n}|
  +1)^{\languidOrder}$, for some constant $C_2>0$ independent of $n$. If we
  now take limits as $n \to \iy$, then $T_n \to \iy$ and $T_{-n} \to -\iy$.
  Since $\languidOrder < 1$, this implies \smash{$|U_n^\integral(\ge)|$} and \smash{$|\genstr_n^\integral(\ge)|$} tend to $0$ as $n \to \iy$.

\subsubsection{Estimating the error term } \label{subsec:proof-error-term}

  To complete the proof of formula \eqref{eqn:tail-term-assertion}, it remains
  to show that the limit $\R_{\ta}(\ge):=\lim_{n\to\iy} \sR_n(\ge)$
  exists and satisfies the asymptotic estimate $\R_{\ta}(\ge)=O(\ge^{d-\sup S})$ as $\ge\to 0^+$, for which we utilize assumption \textbf{\ref{eqn:L2}} of Definition~\ref{def:languid}. Recall from \eqref{error1} and \eqref{eqn:volume-zeta-split2} that the integral $\R_n(\ge)$ is given by
  \linenopax
  \begin{align*}
    \R_n(\ge) &= \frac{1}{2\gp \ii} \int_{S_{|n}} \ge^{d-s} \gzL(s)
                \sum_{k=0}^{d-1} \frac{\genir^{s-k} \crv_k(\gen)(d-k) }{(s-k)(d-s)} ds \\
    &= \frac{1}{2\gp \ii} \int_{T_{-n}}^{T_{n}} \ge^{d-S(t)-\ii t}
       \gzL(S(t)+\ii t) \sum_{k=0}^{d-1}
       \frac{\genir^{S(t)+\ii t-k} \crv_k(\gen)(d-k) }{(S(t)+\ii t-k)(d-S(t)-\ii t)}(S'(t)+i) dt,
  \end{align*}
  where $S'(t)$ denotes the derivative of $S$ at $t$. Note that,
  since $S$ was assumed in Definition~\ref{def:screen} to be Lipschitz continuous with constant
  $\mathrm{Lip} S$, $S'(t)$ exists for almost all $t\in\bR$ and
  $|S'(t)|\le \mathrm{Lip} S$ at those points. Hence the integral
  above is well defined and absolutely integrable, which is seen as follows:
  \linenopax
  \begin{align} \label{eqn:Rn-abs}
    &\frac{1}{2\gp } \int_{T_{-n}}^{T_{n}} \ge^{d-S(t)} |\gzL(S(t)+\ii t)|
      \sum_{k=0}^{d-1} \frac{|\genir^{S(t)+\ii t-k} \crv_k(\gen)|(d-k) }{|S(t)-k +\ii t||d-S(t)+\ii t|}|S'(t)+i| dt \notag \\
    &\hstr[6]\leq \frac{M(\ge)(1+\mathrm{Lip} S)}{2\gp }
      \sum_{k=0}^{d-1} |\crv_k(\gen)|(d-k)   \int_{T_{-n}}^{T_{n}}
      \frac{|\gzL(S(t)+\ii t)| \, dt}{|S(t)-k +\ii t||d-S(t)+\ii t|},
  \end{align}

  where the number $M(\ge)$, defined by
  \linenopax
  \begin{align*}
    M(\ge)
    = \max\{\ge^{d-\sup S}, \ge^{d-\inf S}\} \cdot
      \max\{\genir^{\sup S},\genir^{\inf S}\} \cdot
      \max\{1,\genir^{-d}\},
  \end{align*}
  is a uniform upper bound (in $t$) for the term $\ge^{d-S(t)} \genir^{S(t)-k}$, for $k=0,\dots,d$.
  Now we use the languidness assumption \textbf{\ref{eqn:L2}}, which states
  that there exist constants $C>0$ and $\languidOrder<1$ such that
  $|\gz_s(S(t)+\ii t)|\le C |t|^{\languidOrder}$ for all $|t|\geq 1$.
  Observe that, since the screen $S$ avoids the poles of
  \gzL, the expression $|\gz_s(S(t)+\ii t)|$ is bounded on
  any finite interval for $t$. Therefore, \textbf{\ref{eqn:L2}} is equivalent
  to assuming that there are $C_1>0$ and $\languidOrder<1$ such that
  $|\gz_s(S(t)+\ii t)|\le C_1 |t|^{\languidOrder}$ for all $|t|\geq t_0$,
  where $t_0$ is some arbitrary but fixed positive constant.
  (Simply choose $C_1$ sufficiently large.)

  Next, we describe how to choose $t_0$.
  Since the screen $S$ is assumed to be Lipschitz
  continuous and to avoid the numbers $\{0,\dots,d\}$ when
  passing the real axis, one can find positive constants $t_0$ and $r_0$
  such that $|k-S(t)|\geq r_0$ for all $|t|\le t_0$ and
  $k=0,\dots,d$. (That is, in a tube of width $t_0$ around the
  real axis, the screen $S$ has at least distance $r_0$ to any of
  the lines $\Re s =k$, for $k=0,\dots, d$.)

  Now, for the remaining integrals in the above expression (and
  $n$ sufficiently large), we split the interval of integration
  $(T_{-n},T_n)$ into $(T_{-n}, -t_0)\cup (-t_0, t_0)\cup (t_0,
  T_n)$. In the first and the third intervals, we use (the
  modified) condition \textbf{\ref{eqn:L2}} and, furthermore, that $|d-S(t)+\ii
  t|\geq |t|$ and $|S(t)-k +\ii t|\geq |t|$  to see that, for
  $k=0,1,\dots, d-1$,
  \linenopax
  \begin{align*}
    \int_{T_{-n}}^{-t_0}\frac{|\gzL(S(t)+\ii t)|}{|S(t)-k +\ii t||d-S(t)+\ii t|}  dt
    &\leq C_1 \int_{T_{-n}}^{-t_0} |t|^{\languidOrder-2} dt
    = \frac{C_1}{\languidOrder -1} \left(|T_{-n}|^{\languidOrder-1}-t_0^{\languidOrder-1}\right)
  \end{align*}
  and, similarly, that the \kth integral over the interval
  $(t_0,T_n)$ is bounded by the constant $\frac{C_1}{\languidOrder -1}
  \left(T_{n}^{\languidOrder-1}-t_0^{\languidOrder-1} \right)$.

  In the interval $(-t_0, t_0)$, $|\gzL(S(t)+\ii t)|$ is
  bounded by a constant, say $M$, $|S(t)-k +\ii t|\geq
  |S(t)-k|\geq r_0$ and, similarly,  $|d-S(t)+\ii t|\geq|d-S(t)|\geq r_0$. Therefore, for $k=0,1,\dots,d-1$,
  \linenopax
  \begin{align*}
    \int_{-t_0}^{t_0}\frac{|\gzL(S(t)+\ii t)|}{|S(t)-k +\ii t||d-S(t)+\ii t|} dt
    &\leq \frac{2M t_0}{r_0^{2}}
    =: C_2.
  \end{align*}
  Observe that the derived estimates for the \kth integrals are
  independent of $k$. Thus, putting the pieces back together, we
  have that \eqref{eqn:Rn-abs} is bounded above by
  \linenopax
  \begin{align} \label{eqn:Rn-abs2}

    &C(\ge)  \left(\frac{C_1}{1 - \languidOrder} \left(2 t_0^{\languidOrder-1} - T_n^{\languidOrder-1} - |T_{-n}|^{\languidOrder-1}\right) + C_2\right),
  \end{align}
  where $C(\ge):= \frac{M(\ge)(1+\mathrm{Lip} S)}{2\gp}
  \sum_{k=0}^{d-1} |\crv_k(\gen)|(d-k) $.
  Consequently, $\sR_n(\ge)$ is absolutely integrable for each $n$ and $\ge>0$. Moreover, since \eqref{eqn:Rn-abs2} converges to some finite value as $n \to \iy$ (because $\languidOrder <1$), it follows that also $\sR_{\ta}(\ge)$ is absolutely integrable and thus integrable; i.e., $\sR_{\ta}(\ge)$ is finite for each $\ge>0$. Hence, the error term $\sR_{\ta}(\ge)$ is given as claimed in \eqref{eqn:pointwise-error-head-tail}.
Finally, note that $W_{|n}\to W\cap \{\Re s< c\}$ and
  $\sD_{\tiling}(W)\cap \{\Re s\geq c\}=\{d\}$ imply
  $\sD(W_{|n})\to \sD_{\tiling}(W)\setminus\{d\}$. This completes the proof of
  formula (\ref{eqn:tail-term-assertion}).

  Furthermore, from (\ref{eqn:Rn-abs2}) and the definition of $M(\ge)$ (see the discussion following (\ref{eqn:Rn-abs})), it is clear that there is a constant
  $\hat{C}>0$ such that $|\sR_{\ta}(\ge)|\le \hat{C} \ge^{d-\sup S}$ for
  all $0<\ge<\genir$; i.e., $\sR_{\ta}(\ge)$ is of order $O(\ge^{d-\sup S})$
  as $\ge\to 0^+$.

  Recalling that $\R=\R_{\ta}$, this completes the proof of the languid case in  Theorem~\ref{thm:pointwise-tube-formula}.

  \subsubsection{The strongly languid case}
  \label{subsec:proof-strongly-languid}

  Now assume that \gzL is strongly languid of order $\languidOrder < 2$, as in Definition~\ref{def:strongly-languid} and the second part of Theorem~\ref{thm:pointwise-tube-formula}.
  Then there exists a sequence $S_m$ of screens and corresponding
  windows $W_m$ with $\sup S_m\to -\iy$ as $m\to\iy$ such that
  \textbf{\ref{eqn:L1}} and \textbf{\ref{eqn:L2'}} are satisfied for each $m$ (with
  constants $C,A>0$ independent of $m$). In addition, we may assume without loss of generality that $\sup S_m < S(0)$ for all $m \geq 1$ (see the discussion preceding Corollary~\ref{thm:fractal-spray-tube-formula-monophase}).
  For each screen $S_m$
  and for fixed $n\in\bN$, consider the truncated screen $S_{m|n}$
  (truncated at $T_{-n}$ and $T_n$) and the corresponding
  truncated window $W_{m|n}$ bounded from above and below by the
  horizontal lines $\Im s=T_n$ and $\Im s= T_{-n}$ and from the
  right by the line $\Re s=c$. By the Residue Theorem, for each $m$ and
  $n$, the \nth integral in the sequence of truncated integrals
  in (the counterpart of) \eqref{eqn:V-tail-limit} is given by
  \linenopax
  \[
  \sum_{\pole\in\sD(W_{m|n})} \res[s=\pole]{\gzT[\tail](\ge,s)}+
  \sR_{m|n}(\ge) + U^\integral_{m|n}(\ge) + L^\integral_{m|n}(\ge),
  \]
  just as in the languid case, and the integrals \smash{$U^\integral_{m|n}(\ge)$   and $L^\integral_{m|n}(\ge)$} over the horizontal line segments are similar to \smash{$U_n^\integral(\ge)$} and \smash{$L^\integral_n(\ge)$} above, with $S$ replaced by $S_m$. First we keep $n$ fixed and show that $\sR_{m|n}(\ge)$ vanishes as $m\to\iy$.
  Note that $\sR_{m|n}(\ge)$ is given by the same expression as $\sR_n(\ge)$ in (\ref{error1}), except that the integral is now over $S_{m|n}$ instead of $S_{|n}$. Its absolute value is bounded by
  \linenopax
  \begin{align*} \label{eqn:Rmn-abs}
    \frac{1}{2\gp } \int_{T_{-n}}^{T_{n}} \ge^{d-S_m(t)} |\gzL(S_m(t)+\ii t)|
      \sum_{k=0}^{d-1} \frac{|\genir^{S_m(t)-k +\ii t} \crv_k(\gen)|(d-k) }{|S_m(t)-k +\ii t||d-S_m(t)+\ii t|}|S_m'(t)+i|  dt \\
    \leq \frac{B+1}{2\gp }\sum_{k=0}^{d-1} |\crv_k(\gen)|(d-k)
      \int_{T_{-n}}^{T_{n}} \ge^{d-S_m(t)} \genir^{S_m(t)-k} \frac{|\gzL(S_m(t)+\ii t)|}{|t|^2}  dt,
  \end{align*}
  where we used the inequality $|S_m(t)-k +\ii t||d-S_m(t)+\ii
  t|\geq |t|^2$. Moreover, we utilized that, since the functions
  $S_m$ are assumed to be Lipschitz continuous with a uniform
  Lipschitz bound $B=\sup_m \mathrm{Lip} S_m<\iy$, the inequality
  $|S_m'(t)+i|\le B+1$ holds, whenever $S_m'(t)$ is defined
  (which is the case for almost all $t \in \bR$, independently of $m$).

  Now, by \textbf{\ref{eqn:L2'}} of Definition~\ref{def:strongly-languid}, there are constants $A,C>0$, independent of $n$ and $m$, such that, for
  all $t\in\bR$ and all $m\in\bN$, $|\gzL(S_m(t)+\ii t)|\le C
  A^{|S_m(t)|} (|t|+1)^\languidOrder$. Therefore, there exists a constant
  $C_1$, independent of $n$ and $m$, such that
  \linenopax
  \begin{align*}
    |\sR_{m|n}(\ge)|&\leq C_1
  \int_{T_{-n}}^{T_{n}} \left(\frac\ge\genir\right)^{-S_m(t)}A^{|S_m(t)|}\frac{(|t|+1)^\languidOrder}{|t|^2}
  dt.
  \end{align*}
  For $m$ sufficiently large (indeed, without loss of generality, for all $m \geq 1$), we have $S_m(t)<0$ and so
  $-S_m(t)=|S_m(t)|$. Thus, provided that $\ge < A^{-1} \genir$, we can
  bound the expression $(\ge/\genir)^{-S_m(t)} A^{|S_m(t)|}=(\ge
  A/\genir)^{|S_m(t)|}$ from above by $\left(\ge A/\genir\right)^{|\sup S_m|}$, which is
  independent of $t$ and can thus be taken out of the integral.
  The remaining integral has a finite value for each $n$. Letting
  now $m\to\iy$, $|\sup S_m|\to \iy$ and so $|\sR_{m|n}(\ge)|$
  vanishes.

  When taking the limit as $m\to \iy$, the expression
  \smash{$U^\integral_{m|n}(\ge)$} extends to an integral over the whole half-line
  $(-\iy+\ii T_n, c+\ii T_n]$ and \smash{$L^\integral_{m|n}(\ge)$} to an integral
  over $(-\iy+iT_{-n}, c+\ii T_{-n}]$. More precisely,
  \smash{$U^\integral_{|n}(\ge) :=
  \lim_{m\to\iy} U^\integral_{m|n}(\ge)$} is given by
  \linenopax
  \begin{align*}
    U^\integral_{|n}(\ge)
    &= \frac{1}{2\gp \ii} \int_{-\iy +\ii T_n}^{c+\ii T_n} \gzT[\tail](\ge,s) ds\\
    &= \frac{1}{2\gp \ii} \int_{-\iy}^{c} \ge^{d-t-\ii T_n} \gzL(t+\ii T_n)
      \sum_{k=0}^{d-1} \frac{\genir^{t-k+\ii T_n} \crv_k(\gen)(d-k) }{(t-k+\ii T_n)(d-t-\ii T_n)} dt.
  \end{align*}
  By exploiting the languidness condition \textbf{\ref{eqn:L1}} (which now
  holds for all $t \in \bR$) and the inequalities  $|t-k+\ii T_n| \geq T_n$
  (for $k=0,\dots,d-1$) and $|d-t-\ii T_n| \geq T_n$, it is easily
  seen that there exists some constant $C_2 > 0$, independent of $n$ and $m$, such that $U^\integral_{|n}(\ge)$ is absolutely bounded as follows:
  \linenopax
  \begin{equation} \label{eqn:u-n-L2}
    \left|U^\integral_{|n}(\ge) \right|
    \leq C_2 \frac{(T_n +1)^\languidOrder}{T_n^2} \int_{-\iy}^{c} \left(\frac\ge\genir\right)^{-t} dt.

  \end{equation}
  The remaining integral is finite, provided that $\ge < \genir$. Now,
  as $n \to \iy$, \smash{$|U^\integral_{|n}(\ge)|$} vanishes, for each $\ge < \genir$, and
  with completely analogous arguments, the same can be shown for
  \smash{$|U^\integral_{|n}(\ge)|$}. Hence the tail volume is given in the strongly languid case by
  \linenopax
  \begin{equation}
    V_{\mathrm{tail}}(\tiling,\ge)
    = \sum_{\pole\in\sD_{\tiling}} \res[s=\pole]{\gzT[\tail](\ge,s)};
  \end{equation}
  i.e., equation \eqref{eqn:tail-term-assertion} holds without error term for $\ge<\min\{\genir, A^{-1} \genir\}$.
  This completes the proof for the strongly languid case, and thus of all of Theorem~\ref{thm:pointwise-tube-formula}.

\subsubsection{Proof of Corollary~\ref{thm:fractal-spray-tube-formula-monophase}}
\label{sec:proof-of-cor-monophase}
  In the monophase case, we have $f_k(\ge) = 0$ for all $\ge > 0$ and for each $k=0,1,\dots,d$. Therefore, $\gzT[\head]$ (given in \eqref{eqn:gzThead}) vanishes identically, implying $\R_{\he}\equiv 0$ (by \eqref{eqn:pointwise-error-head}) and thus $V_{\mathrm{head}}(\tiling,\ge)=0$ for each $\ge>0$ (by~\eqref{eqn:head-term-assertion}). Consequently, by \eqref{eqn:def:V-tail}, $V(\tiling,\mydot)=V_{\mathrm{tail}}(\tiling,\mydot)$.  Since $\R_{\ta}(\ge)=\R(\ge)$, cf.~\eqref{eqn:pointwise-error} and \eqref{eqn:pointwise-error-head-tail}, the assertion of Corollary~\ref{thm:fractal-spray-tube-formula-monophase} follows by observing that the assumption $S(0)<0$ is not used in the proof of \eqref{eqn:tail-term-assertion} given in Section~ \ref{subsec:proof-tail} and Section~ \ref{subsec:proof-error-term}. It is only used to ensure that the screen $S$ avoids the integer dimensions $0,1,\ldots,d$. Note that $\gzT[\head]\equiv 0$ also implies $\gzT=\gzT[\tail]$. Hence, the error term $\R(\ge)$ is equivalently given by the integral \eqref{eqn:pointwise-error2} in this case, as explained in Remark~\ref{rem:error-term}.

\hfill $\square$

\subsection{Proof of the fractal tube formula, Corollary~\ref{thm:ptwise-result-self-similar-case-simplified}}
\label{sec:proof-of-fractal-tube-formula}
\hfill \\

Before proceeding, we need to compute some residues. To this end, we introduce the tubular zeta function for the generator \gen. In addition to being a useful technical device, it reveals the structure of the residues of the tubular zeta function \gzT.

\begin{defn}\label{def:gzG}
  Let $\gzG(\ge,s)$ denote the \emph{tubular zeta function of the generator} \gen, where \gen is assumed to have a Steiner-like representation as in \eqref{eqn:def-prelim-Steiner-like-formula}.
It is defined exactly as $\gzT(\ge,s)$, except that the associated fractal string is given by $\{\genstr_j\}_{j=1}^\iy$ with $\genstr_1 = 1$ and $\genstr_j = 0$ for all $j \geq 2$. In other words, it is the tubular zeta function of the trivial fractal spray with generator \gen.
\end{defn}

Exactly as in \eqref{eqn:volume-zeta-split2}, we write $\gzG = \gzG[\head] + \gzG[\tail]$, so that for $s \in \bC$, we have
  \linenopax
  \begin{align}\label{eqn:gzG[head]}
    \gzG[\head](\ge,s) =
    \begin{cases}
      \ge^{d-s} \sum_{k=0}^d \frac{\genir^{s-k}}{s-k} f_k(\ge),
      & 0 < \ge \leq \genir, \\
      0, & \ge \geq \genir,
    \end{cases}
  \end{align}
  with $f_k(\ge) = \crv_k(\gen,\ge) - \crv_k(\gen)$ defined as in \eqref{eqn:def:f_k} for $k=0,1,\dots,d$, and
  \linenopax
  \begin{align}\label{eqn:gzG[tail]}
    \gzG[\tail](\ge,s) = \frac{\ge^{d-s}}{d-s} M_s(\gen),
    \qq \ge > 0,
  \end{align}
  where
  \linenopax
  \begin{align}\label{eqn:M_s(G)}
    M_s(\gen) := \sum_{k=0}^{d-1} \frac{\genir^{s-k}}{s-k} (d-k)\crv_k(\gen).
  \end{align}

  To see why the second case of \eqref{eqn:gzG[head]} should be true, consider the definition of \gzG[\head] in the counterpart of \eqref{eqn:volume-zeta-split2} and suppose we define
  \linenopax
  \begin{align}\label{eqn:J_G(e)}

    J_G(\ge) := \charfn{(0,\genir)}(\ge),
  \end{align}
  which is in parallel to \eqref{eqn:J(eps)}, upon inspection.

  Observe that for every $\ge > 0$, $\gzG(\ge,\mydot)$ is meromorphic in all of \bC, with poles contained in $\{0,1,\dots,d\}$. Hence, the set of ``complex dimensions'' of \gen consists of the integer dimensions $\{0,1,\dots,d\}$, and all of these poles are simple.

\begin{lemma}[Residues of \gzG]
  \label{thm:gzG-residues}
  For $0 < \ge \leq \genir$, we have the following residues of \gzG:
  \linenopax
  \begin{align}
    &\res[s=k]{\gzG[\head](\ge,s)} = \ge^{d-k} f_k(\ge),
      && k=0,1,\dots,d,
      \label{eqn:gzG[head]-residue-at-k} \\
    &\res[s=k]{\gzG[\tail](\ge,s)} = \ge^{d-k} \crv_k(\gen),
      && k=0,1,\dots,d-1,
      \label{eqn:gzG[tail]-residue-at-k} \\
    &\res[s=k]{\gzG[\tail](\ge,s)} = \crv_d(\gen) - \gl_d(\gen),&& k=d.
      \label{eqn:gzG[tail]-residue-at-d}
  \end{align}
  \begin{proof}
    In light of \eqref{eqn:gzG[head]}--\eqref{eqn:M_s(G)}, each of $\gzG[\head](\ge,\mydot)$, $\gzG[\tail](\ge,\mydot)$ and $\gzG(\ge,\mydot)$ is meromorphic in all of \bC, with (simple) poles contained in $\{0,1,\dots,d\}$, for $0 < \ge \leq \genir$. To show \eqref{eqn:gzG[tail]-residue-at-d}, simply use \eqref{eqn:gzG[tail]} and \eqref{eqn:M_s(G)} to compute
    \linenopax
    \begin{align*}
      \res[s=d]{\gzG[\tail](\ge,s)}
      &= \lim_{s \to d} \, (s-d) \gzG[\tail](\ge,s)
      = - \sum_{k=0}^{d-1} \genir^{d-k} \crv_k(\gen)

      = \crv_d(\gen) - \gl_d(\gen),
    \end{align*}
    using \eqref{eqn:leb-coeff-rel} to reach the last equality.
  \end{proof}
\end{lemma}

The following result will not be used in the sequel but may be helpful for the reader; it provides a ``residue formulation'' of the given Steiner-like representation of \gen.

\begin{comment}
  \label{thm:Exact-tube-formula-for-G}
  For all $\ge > 0$, we have
  \begin{align}\label{eqn:residue-tube-formula-for-G}
    V(\gen,\ge) = \sum_{k=0}^d \res[s=k]{\gzG(\ge,s)} + \gl_d(\gen).
  \end{align}
  \begin{proof}
    This follows from a direct computation by using \eqref{eqn:gzG[head]-residue-at-k}--\eqref{eqn:gzG[tail]-residue-at-d}, along with \eqref{eqn:def-prelim-Steiner-like-formula} and \eqref{eqn:leb-coeff-rel}. That is, one verifies that the sum of the residues of $\gzG[\head](\ge,s)$ (resp. of $\gzG[\tail](\ge,s)$) at the integer dimensions $s=k \in\{0,1,\dots,d\}$ is equal to $V(\gen,\ge) - \gl_d(\gen)$ (resp. $0$). The first sum is initially computed for $0 < \ge < \genir$ using \eqref{eqn:def:f_k} and \eqref{eqn:gzG[head]-residue-at-k}, and is then seen to also hold for $\ge \geq \genir$, since $\gzG[\head] \equiv 0$ for these values of \ge. Formula \eqref{eqn:residue-tube-formula-for-G} then follows by adding these two sums.
  \end{proof}
\begin{remark}\label{rem:residue-tube-formula-for-G}
  Corollary~\ref{thm:Exact-tube-formula-for-G} follows from our main result, Theorem~\ref{thm:pointwise-tube-formula}, but it is enlightening to verify it directly. This result will not be used in the sequel, but may serve as a helpful guide to the reader.
\end{remark}
\end{comment}

\begin{cor}[Exact tube formula for \gen]\label{thm:Exact-tube-formula-for-G}
  For all $\ge \in (0,\genir]$,
  \linenopax
  \begin{align}\label{eqn:residue-tube-formula-for-G}
    V(\gen,\ge) = \sum_{k=0}^d \res[s=k]{\gzG(\ge,s)} + \gl_d(\gen).
  \end{align}
  \begin{proof}
    First, note that it follows from \eqref{eqn:gzG[head]-residue-at-k} and \eqref{eqn:def:f_k} that for $0 < \ge \leq \genir$,
    \linenopax
    \begin{align}
      \sum_{k=0}^d \res[s=k]{\gzG[\head](\ge,s)}
      &= \sum_{k=0}^d \ge^{d-k} \left(\crv_k(\gen,\ge) - \crv_k(\gen)\right) \notag \\
      &= \sum_{k=0}^d \ge^{d-k} \crv_k(\gen,\ge) - \sum_{k=0}^d \ge^{d-k} \crv_k(\gen) \notag \\
      &= V(\gen,\ge) - \sum_{k=0}^{d-1} \ge^{d-k} \crv_k(\gen) - \crv_d(\gen),
      \label{eqn:residue-tube-formula-for-G-derivation1}
    \end{align}
    where we have used \eqref{eqn:def-prelim-Steiner-like-formula} in the last equality.

    Furthermore, by \eqref{eqn:gzG[tail]-residue-at-k} and \eqref{eqn:gzG[tail]-residue-at-d}, we have for $\ge \in (0,\genir]$,
    \linenopax
    \begin{align}
      \sum_{k=0}^d \res[s=k]{\gzG[\tail](\ge,s)}
      = \sum_{k=0}^{d-1} \ge^{d-k} \crv_k(\gen) + \left(\crv_d(\gen)- \gl_d(\gen)\right).
      \label{eqn:residue-tube-formula-for-G-derivation2}
    \end{align}
    Since $\gzG = \gzG[\head] + \gzG[\tail]$, the result now follows by adding \eqref{eqn:residue-tube-formula-for-G-derivation1} and \eqref{eqn:residue-tube-formula-for-G-derivation2}.
  \end{proof}
\end{cor}

As an alternative proof of Corollary~\ref{thm:Exact-tube-formula-for-G}, one can  obtain \eqref{eqn:residue-tube-formula-for-G} by applying the second part of Theorem~\ref{thm:pointwise-tube-formula} to the trivial fractal spray on \gen. However, we feel that the proof given above is more edifying and more straightforward.

\subsubsection{The residues of \gzT}
\label{sec:residues-of-gzT}

Let \tiling be a self-similar tiling with a fractal string $\sL = \{\scale_j\}_{j=1}^\iy$ and a single generator \gen for which a Steiner-like representation has been fixed. Let $\gzT = \gzT(\ge,s)$ denote the tubular zeta function of \tiling, and let $\gzT = \gzT[\head] + \gzT[\tail]$ be its head--tail decomposition, as in Section~\ref{sec:splitting}. In light of \eqref{eqn:volume-zeta-split2}, we deduce from \eqref{eqn:gzG[tail]} and \eqref{eqn:M_s(G)} that $\gzT[\tail]$ factors as follows:
  \linenopax
  \begin{align}\label{eqn:gzT[tail]-factored}
    \gzT[\tail](\ge,s) = \gzG[\tail](\ge,s) \gzL(s).
  \end{align}
Furthermore, still by \eqref{eqn:volume-zeta-split2},
  \linenopax
  \begin{align}\label{eqn:gzT[head]-notfactored}
    \gzT[\head](\ge,s)
    = \ge^{d-s} \sum_{k=0}^d \frac{\genir^{s-k}}{s-k}
      \sum_{j=1}^{J(\ge)} \scale_j^s f_k(\scale_j^{-1} \ge),
  \end{align}
  with $f_k$ as in \eqref{eqn:def:f_k} and $J(\ge)$ as in \eqref{eqn:J(eps)}, as usual.
  Recall that $J(\ge) \to \iy$ monotonically as $\ge \to 0^+$, since $\scale_j$ decreases monotonically to $0$ as $j \to \iy$.

\begin{remark}\label{rem:gzT[head]-notfactored}
  Observe that \eqref{eqn:gzT[head]-notfactored} is not at all the counterpart of the factorization given in \eqref{eqn:gzT[tail]-factored}. Indeed, it clearly does not enable us to write $\gzT[\head]$ as the product of $\gzG[\head]$ and \gzL (which would be false). This is the source of some difficulty if we wish to estimate the residues of $\gzT[\head](\ge,s)$ as $\ge \to 0^+$.

\end{remark}

\begin{lemma}[Residues of \gzT]
  \label{thm:residues-of-gzT}
  Fix $\ge \in (0,\gen]$. Then:
  \begin{enumerate}
    \item[\emph{(i)}] When $\pole \in \sD_\sL \less \{0,1,\dots,d\}$ is a simple pole of \gzL, the residue $\res{\gzT(\ge,s)}$ is given by
      \linenopax
      \begin{align}\label{eqn:residues-of-gzT(i)}
        \gzG[\tail](\ge,\pole) \res{\gzL(s)}
        = \frac{\ge^{d-\pole}}{d-\pole} \res{\gzL(s)} M_\pole(\gen),
      \end{align}
      with $M_\pole(\gen) = \sum_{k=0}^{d-1} \frac{\genir^{\pole-k}}{\pole-k}(d-k)\crv_k(\gen)$ as in \eqref{eqn:M_s(G)}.
    \item[\emph{(ii)}] For $\pole = k \in \{0,1,\dots,d\} \less \sD_\sL$,
    \footnote{\label{foot}Note that if \tiling is a self-similar tiling, then the only real pole of \gzL is $\abscissa = \abscissa_\sL < d$. Hence, the only way that \pole could belong to both $\sD_\sL$ and $\{0,1,\dots,d\}$ would be if $\pole = \abscissa = k$, for some $k \in \{0,1,\dots,d-1\}$.}
    the residue $\res[s=k]{\gzT[\tail](\ge,s)}$ is given by
      \linenopax
      \begin{align}\label{eqn:residues-of-gzT(ii)}
        \res[s=k]{\gzT[\tail](\ge,s)}
        = \begin{cases}
          \ge^{d-k} \crv_k(\gen) \gzL(k), & k \neq d, \\
          \left(\crv_d(\gen) - \gl_d(\gen) \right) \gzL(d), &k=d.
        \end{cases}
      \end{align}

    \item[\emph{(iii)}] For $\pole = k \in \{0,1,\dots,d\}$, the residue $\res[s=k]{\gzT[\head](\ge,s)}$ is given by
      \linenopax
      \begin{align}\label{eqn:residues-of-gzT(iii)}
        \res[s=k]{\gzT[\head](\ge,s)}
        = \ge^{d-k} \sum_{j=1}^{J(\ge)} \scale_j^k f_k(\scale_j^{-1} \ge),
      \end{align}
      with $J(\ge)$ as in \eqref{eqn:J(eps)} and $f_k$ as in \eqref{eqn:def:f_k}.

  \end{enumerate}
  \begin{proof}
    In light of the factorization formula \eqref{eqn:gzT[tail]-factored},
    \eqref{eqn:residues-of-gzT(i)} follows from \eqref{eqn:gzG[tail]} and the fact that, under the assumption of (i),
    \linenopax
    \begin{align*}
      \res{\gzT(\ge,s)} &= \res{\gzT[\tail](\ge,s)},
      \qq \text{for } \pole \in \DL \less \{0,1,\dots,d\},
    \end{align*}
    while \eqref{eqn:residues-of-gzT(ii)} follows from \eqref{eqn:gzG[tail]-residue-at-k}--\eqref{eqn:gzG[tail]-residue-at-d} of Lemma~\ref{thm:gzG-residues}. Note that \pole is a simple pole of \gzL in part (i), and hence it is at most a simple pole of $\gzT[\tail](\ge,s)$; whence
    \linenopax
    \begin{align}\label{eqn:gzTtail-res-at-w}
      \res{\gzT[\tail](\ge,s)}
      = \lim_{s \to \pole} (s-\pole) \gzT[\tail](\ge,s)

      = \gzG[\tail](\ge,\pole) \res{\gzL(s)},
    \end{align}
    from which \eqref{eqn:residues-of-gzT(i)} follows in light of \eqref{eqn:gzG[tail]}.
    Since $\pole = k \in \{0,1,\dots,d\}$ is a simple pole of $\gzT[\head]$,
    \linenopax
    \begin{align*}
      \res[s=k]{\gzT[\head](\ge,s)} = \lim_{s \to k} \, (s-k) \gzT[\head](\ge,s),
    \end{align*}
    and hence \eqref{eqn:residues-of-gzT(iii)} follows immediately from \eqref{eqn:gzT[head]-notfactored}.

    Finally, as was already observed, the poles of $\gzT[\head]$ and $\gzT[\tail]$ belong to $\{0,1,\dots,d\}$ and $\sD_\tiling$, respectively.
  \end{proof}
\end{lemma}

\begin{remark} \label{rem:residues-of-gzT}
Note that Lemma~\ref{thm:residues-of-gzT} is valid for an arbitrary  fractal spray satisfying the hypotheses of the first part of Theorem~\ref{thm:pointwise-tube-formula}, but without the assumption that $S(0)<0$ (which is not necessary for Lemma~\ref{thm:residues-of-gzT} to hold).
\end{remark}

\subsubsection{The proof of Corollary~\ref{thm:ptwise-result-self-similar-case-simplified}}
  \label{sec:proof-of-some-corollary}
  Let \tiling be a self-similar tiling satisfying the hypotheses of Corollary~\ref{thm:ptwise-result-self-similar-case-simplified}.
  Note that since the poles of \gzT are assumed to be simple, it follows that $\sD_\sL$ and $\{0,1,\dots,d\}$ are disjoint; that is, all poles of \gzL are simple and $D \notin \{1,\dots,d-1\}$.
Recall that since \tiling is a self-similar tiling, we have $0 < \abscissa < d$ and \abscissa is the only pole of \gzL on the real axis. See footnote 6.
The following proof makes use of the decomposition $\gzT = \gzT[\head] + \gzT[\tail]$ from \eqref{eqn:volume-zeta-split2}.

    Since $\sD_\tiling = \sD_\sL \cup \{0,1,\dots,d\}$ is a disjoint union, the present hypotheses and Theorem~\ref{thm:ptwise-result-self-similar-case} yield
    (for $\ge\in(0,g)$ and with $e_k(\ge)$ defined as in \eqref{eqn:self-similar-pointwise-tube-formula-coefficients-ek}, for $k=0,1,\dots,d$)
    \linenopax
    \begin{align}
    V(\tiling,\ge)

      &= \sum_{\pole \in \sD_\sL} \res[\pole]{\gzT[\tail](\ge,s)}
       + \sum_{k \in \{0,1,\dots,d\}} \res[k]{\gzT[\tail](\ge,s)}
       \notag \\
       &\hstr[15]+ \sum_{k \in \{0,1,\dots,d\}} \res[k]{\gzT[\head](\ge,s)} + \gl_d(\gen) \gzL(d)
       \label{eqn:fractal-tube-formula-derivation1} \\
      &= \sum_{\pole \in \sD_\sL} \frac{\ge^{d-\pole}}{d-\pole} M_\pole(\gen) \res[\pole]{\gzL(s)}
       + \sum_{k=0}^{d-1} \ge^{d-k} \crv_k(\gen) \gzL(k)
       + \left(\crv_d(\gen) - \gl_d(\gen)\right)\gzL(d) \notag \\
       &\hstr[15]+ \sum_{k=0}^d \ge^{d-k} e_k(\ge) + \gl_d(\gen)\gzL(d),
       \label{eqn:fractal-tube-formula-derivation2}
    \end{align}
    from which \eqref{eqn:self-similar-pointwise-tube-formula-simplified} follows.

    In \eqref{eqn:fractal-tube-formula-derivation2}, we have set
    \linenopax
    \begin{align*}
      M_\pole(\gen) = \sum_{k=0}^{d-1} \frac{\genir^{\pole-k}}{\pole-k}(d-k) \crv_k(\gen)
    \end{align*}
    as in \eqref{eqn:M_s(G)}, so that
    $c_\pole = M_\pole(\gen) \res{\gzL(s)}/(d-\pole)$ for $\pole \in \sD_\sL$, and applied Lemma~\ref{thm:residues-of-gzT} to obtain the precise values of the residues of $\gzT[\head]$ and $\gzT[\tail]$;
    see \eqref{eqn:residues-of-gzT(i)}--\eqref{eqn:residues-of-gzT(iii)}.
    In particular, this verifies \eqref{eqn:self-similar-pointwise-tube-formula-coefficients-cw}.
    Note also that the residue of $\gzT[\tail](\ge,s)$ at $s=d$ and the term $\gl_d(\gen) \gzL(d)$ have been combined to yield $\crv_d(\gen) \gzL(d) = c_d$. This verifies \eqref{eqn:self-similar-pointwise-tube-formula-coefficients-ck}.

    Note that the expression of $c_k$ given in \eqref{eqn:self-similar-pointwise-tube-formula-coefficients-ck} for $k \in \{0,1,\dots,d-1\}$ follows from the second sum in \eqref{eqn:fractal-tube-formula-derivation2}.

\begin{comment}\label{rem:H_k-in-fractal-tube-formula}
  Given $k \in \{0,1,\dots,d\}$ with $k > D$, Appendix~\ref{sec:residues-of-gzT} gives a natural condition (hypothesis \ref{eqn:H_k}, stated just before Lemma~\ref{thm:residue-decay-rate}, under which $e_k(\ge) \to 0$ as $\ge \to 0^+$, and hence $e_k(\ge)$ in \eqref{eqn:self-similar-pointwise-tube-formula-coefficients-ek} can be viewed as a small perturbation of the constant $c_k$ appearing in \eqref{eqn:self-similar-pointwise-tube-formula-coefficients-ck}. Note also that $k > D$ implies $\gzL(k) > 0$. Furthermore, it follows from Appendix~\ref{sec:residues-of-gzT} that if there exists a $\gt_k$ such that $0 < \gt_k < k-D$ and
  \linenopax
  \begin{align}\label{eqn:}
    \crv_k(\gen,\ge) = \crv_k(\gen) + O(\ge^{\gt_k}),
    \qq \text{as } \ge \to 0^+,
  \end{align}
  then $e_k(\ge) = O(\ge^{\gt_k})$, and hence $e_k(\ge) \to 0$ as $\ge \to 0^+$. It is important to keep this in mind when applying Corollary~\ref{thm:ptwise-result-self-similar-case-simplified} or, more generally, Theorem~\ref{thm:pointwise-tube-formula}. (Note that since $k > D$, it is necessarily the case that $k$ is a simple pole of \gzT because then $k \notin \sD_\sL$.) This comment is used in the discussion of Minkowski measurability in Section~\ref{sec:Minkowski-measurability}, below.
\end{comment}

config: arxiv


TITLE: Smallest axis-aligned bounding box of hyper-ellipsoid
QUESTION [4 upvotes]: Let $E$ be the $n$-dimensional ellipsoid defined by
$$E:=\{x \in \mathbb{R}^n: (x-c)^T A (x-c) \le 1\},$$
where $c \in \mathbb{R}^n$ is the center of the ellipsoid, and $A \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix.
Question:
How can one efficiently compute the coordinate-axis-aligned bounding box that just barely contains the ellipsoid?
For a 2D example, see the following picture:

Note:
I ask this question, and answer it myself, because this question (in general form) is surprisingly absent from math.stackexchange even after 10+ years. Good answers to this question are hard to find on the internet more generally. After googling around, I eventually had to figure this out myself, and am posting here to spare future people the same trouble. Many websites discuss the question in the special case of $2D$ and $3D$, but the the format of the ellipse is given in terms of axes and angles rather than SPD matrices, and the formulas do not generalize to n-dimensions. The good answer is given by achilles hui in the comments to the following closed question:
Bounding box of Ellipsoid
but no proof is provided there, and the question is closed so I cannot provide the answer with proof there. Even if that question were reopened, it is focused on the 3D case with axes and angles rather than the n-dimensional case with SPD matrices.

REPLY [4 votes]: Given vector $\rm{c} \in \Bbb R^n$ and matrix $\rm{Q} \succ \rm{O}_n$, let
$$\mathcal E := \left\{ \rm{x} \in \Bbb R^n \mid \left( \rm{x} - \rm{c} \right)^\top \rm{Q}^{-1} \left( \rm{x} - \rm{c} \right) \leq 1 \right\}$$
Let $g (\rm{x}) := \left( \rm{x} - \rm{c} \right)^\top \rm{Q}^{-1} \left( \rm{x} - \rm{c} \right)$. The vector field orthogonal to the boundary of ellipsoid $\mathcal E$ is
$$\nabla g (\rm{x}) = 2 \, \rm{Q}^{-1} \left( \rm{x} - \rm{c} \right)$$
Let us pick $i \in [n]$ and focus on the $i$-th axis. Let $\rm{P}_i := \rm{e}_i \rm{e}_i^\top$ be the projection matrix that projects onto the $i$-th axis. At the two points where ellipsoid $\mathcal E$ touches the (smallest) bounding box, we have $\rm{P}_i \nabla g (\rm{x}) = \nabla g (\rm{x})$, i.e.,
$$\left( \rm{I}_n - \rm{P}_i \right) \underbrace{ {\rm Q}^{-1} \left( \rm{x} - \rm{c} \right)}_{=: {\rm y}} = 0_n$$
Hence, $y_i$ is free and all other entries of $\rm y$ are zero, i.e., ${\rm y} = t \, {\rm e}_i$, or, ${\rm x} = {\rm c} + t \, {\rm Q} \, {\rm e}_i$. Intersecting this line with the boundary of ellipsoid $\mathcal E$, we obtain
$$t^2 = \left( {\rm e}_i^\top {\rm Q} \, {\rm e}_i \right)^{-1} = q_{ii}^{-1}$$
or, $t = \pm \frac{1}{\sqrt{q_{ii}}}$. Thus, ellipsoid $\mathcal E$ touches the (smallest) bounding box at points
$${\rm x} = {\rm c} + t \, {\rm Q} \, {\rm e}_i = {\rm c} \pm \frac{1}{\sqrt{q_{ii}}} \, {\rm Q} \, {\rm e}_i$$
and, projecting onto the $i$-th axis,
$$x_i = c_i \pm \frac{1}{\sqrt{q_{ii}}} \, {\rm e}_i^\top {\rm Q} \, {\rm e}_i = c_i \pm \frac{q_{ii}}{\sqrt{q_{ii}}} = c_i \pm \sqrt{q_{ii}}$$
Hence, the bounding box is
$$\color{blue}{\left[ c_1 - \sqrt{q_{11}}, c_1 + \sqrt{q_{11}} \right] \times \left[ c_2 - \sqrt{q_{22}}, c_2 + \sqrt{q_{22}} \right] \times \cdots \times \left[ c_n - \sqrt{q_{nn}}, c_n + \sqrt{q_{nn}} \right]}$$


TITLE: Condition for vector to be independent
QUESTION [0 upvotes]: In the following picture the "BUTTERFLY NETWORKS" with edge capacity  equal $1$,

we define vector associated  to edge $ST$ by $f_{ST}$, so we have the vectors for all edges as :

and also from the max flow min cut theorem, the max flow from source $S$ to any node in the network is the number of edges whose removable disconnect $S$ to this node , because the capacity of edges here is one.
we not the max flow from $S$ to any node non source (not $S$)  $T$ on this network by $$maxflow(T).$$For example $maxflow(U)=1$,$maxflow(Z)=2$.
Write $V_T =\langle \{f_e : e \in In(T)\} \rangle$ , where $\langle \cdot \rangle$ is spane, and $e \in In(T)$ the vector for edges in in node $T$.
For example $V_W =\langle \{f_{TW},f_{UW} \} \rangle$.
We define for this example $\omega=2$.
and we have the following condition :
(2.5)  $dim(V_T) = \omega$ for every non-source ( not node $S$) node $T$ with $maxflow(T) ≥ ω$.
So to meets the criterion (2.5) for this example we must have:
$f_{TW}$ and $f_{UW}$ are linearly independent.
$f_{TY}$ and $f_{XY}$ are linearly independent.
$f_{UZ}$ and $f_{XZ}$ are linearly independent.
Equivalently, the criterion says that $$s,t,u,v,y,z, nr − pq$$
$$npsw + nrux - pnsw - pqux, \text{ and } rnsw + rqux - qpsw - qrux$$ are all nonzero.
My question is, I haven't understood why :
Equivalently, the criterion says that
 $$s,t,u,v,y,z, nr − pq$$ $$npsw + nrux - pnsw - pqux, \text{ and } rnsw + rqux - qpsw - qrux$$ are all nonzero ?.
I hope to find the answer, thanks.

REPLY [1 votes]: First, each vector in the constraint cannot be a zero vector. For example, if $s=0$ then $f_{TW}$ becomes a zero vector. Therefore, $s \neq 0$. Likewise, you can prove $t,u,v,y,z$ are also non-zero.
Second, two vectors $v_1 = \begin{bmatrix}\alpha \\ \beta \end{bmatrix}$ and $v_2 =\begin{bmatrix}\gamma \\ \delta \end{bmatrix}$ are linearly independent when $$\frac{\alpha}{\beta} \neq \frac{\gamma}{\delta}.$$ Equivalently, when $$\alpha \delta - \beta \gamma\neq 0.$$ Using this relationship, it is easy to verify that other expressions should not be equal to zero.