TITLE: The inequality $\int^\infty_0 \frac{\sin(rt)}{rt}\frac{r^4}{\sinh^2(r)} e^{-ar\coth(r)}dr\leq c \big(e^{-At}\big)$
QUESTION [2 upvotes]: Let $a>0$. How to prove the following inequality $$\exists c>0,\exists A>0,\forall t>0:\quad\int^\infty_0 \frac{\sin(rt)}{rt}\frac{r^4}{\sinh^2(r)} e^{-ar\coth(r)}dr\leq c \big(e^{-At}\big)$$
for some constants $A>0,c>0$

REPLY [3 votes]: $\newcommand{\R}{\mathbb R}\renewcommand{\S}{\mathcal S}$This follows from Theorem IX.14, which states the following:

Let $T\in\S'(\R^n)$. Suppose that the Fourier transform $\hat T$ of $T$ can be continued analytically to the set $\{z\colon|\Im z|<A\}$ for some $A>0$. Suppose also that for each $B<A$ we have $\sup\{\|\hat T(\cdot+iy)\|_1\colon y\in(-B,B)\}<\infty$. Then $T$ is a bounded continuous function and for each $B<A$ there is a real number $C_B$ such that
\begin{equation*}
    |T(x)|\le C_B e^{-B|x|}
\end{equation*}
for $x\in\R^n$.

Indeed, for the integral in question,
\begin{equation*}
    I(t):=\int^\infty_0 \frac{\sin rt}{rt}\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr,
\end{equation*}
and
\begin{equation*}
    T(t):=t I(t), \tag{1}\label{1}
\end{equation*}
we have $T=\frac{\sqrt{2\pi}}{2i}\,\check f$, where
\begin{equation*}
    f(r):=\frac{r^3}{\sinh^2r} e^{-ar\coth(r)}
\end{equation*}
and $\check f$ is the inverse Fourier transform of $f$ (defined by the formula $\check f(t):=\frac1{\sqrt{2\pi}}\,\int_\R e^{irt}f(r)\,dr$), so that
\begin{equation*}
    \hat T=\frac{\sqrt{2\pi}}{2i}\,f.
\end{equation*}
It is not hard to see that the function $T$ defined by \eqref{1} satisfies all the conditions of Theorem IX.14, cited above, with $A=\pi$. So, for each $B\in(0,\pi)$ there is a real number $C_B$ such that $|T(t)|\le C_B e^{-Bt}$ for all $t\ge0$ and hence
\begin{equation*}
    |I(t)|\le C_B e^{-Bt}
\end{equation*}
for $t\ge1$. On the other hand, for all $t\in[0,1]$,
\begin{equation*}
    |I(t)|\le c_a:=\int^\infty_0\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr
    \le c_{a,B}\, e^{-Bt},
\end{equation*}
where $c_{a,B}:=c_a e^B\in(0,\infty)$.
Thus, for each $B\in(0,\pi)$ and all $t\ge0$
\begin{equation*}
    |I(t)|\le\max(C_B,c_{a,B}) e^{-Bt},
\end{equation*}
as desired.


TITLE: What is the non-recursive sigma notation to describe $k$-th derivative of a polynomial function?
QUESTION [0 upvotes]: I need to to determine the sigma notation to describe $k$-th derivative of  a polynomial function.
For example if the polynomial function is:
$$
f(x)=a_0+a_1x^1+a_2x^2+a_3x^3+a_4x^4
$$
then then if $k=2$ the $k$-th derivative is:
$$
f''(x)=2a_2+3!a_3x+3\cdot 4a_4x^2
$$
I know there're recursive definitions of $k$-th derivative but I need to determine a non-recursive formula.
In order to generalize the sum of $f^{(k)}$ I came up with this:
$$
f^{(k)}=k!a_k+\sum_{i=k+1\\j=1}^n \frac{i!}{j!}a_ix^j
$$
Is there an established formula for this?

REPLY [2 votes]: You basically have the good formula but it is expressed strangely with a mix of $i$ and $j$'s.
$p(x)=\sum\limits_{i=0}^n a_i\,x^i$
$p'(x)=\sum\limits_{i=1}^n ia_i\,x^{i-1}\quad$ we start at $i=1$ since $a_0$ has disappeared.
$p''(x)=\sum\limits_{i=2}^n i(i-1)a_i\,x^{i-2}\quad$ starting at $i=2$.
$p^{(k)}(x)=\sum\limits_{i=k}^n i(i-1)\cdots(i-k+1)a_i\,x^{i-k}\quad$ starting at $i=k$.
We can re-index the sum to have powers of $x$ starting at $0$ (set $j=i-k$)
$\displaystyle p^{(k)}(x)=\sum\limits_{j=0}^{n-k} (j+1)(j+2)\cdots (j+k)a_{j+k}\,x^j=\sum\limits_{j=0}^{n-k} \dfrac{(j+k)!}{j!}a_{j+k}\,x^j$
Another possibility is to factorize $k!$ to make binomial coefficient appear.
$\displaystyle \dfrac 1{k!}\,p^{(k)}(x)=\sum\limits_{j=0}^{n-k} C_{j+k}^k\,a_{j+k}\,x^j$

REPLY [2 votes]: notice for $f=ax^n$ we have $f^{(k)}=a\frac{n!}{(n-k)!}x^{n-k}$
you can think of $\frac{n!}{(n-k)!}$ as factorial in the opposite direction, instead of $1\cdot2\cdot\cdots\cdot k$ we have $n\cdot (n-1)\cdot(n-2)\cdot\cdots \cdot(n-k+1)$.
now for $f=a_1x+a_2x^2+\cdots$ the only thing that changes is $n$, so we get $$f^{(k)}=\sum_{j=0}^{n}a_j\frac{j!}{(j-k)!}x^{j-k}$$
but notice that when $j<k$ we get $\frac{j!}{(j-k)!}=0$, so we can start from $j=k$ to get the final result$$\boxed{f^{(k)}=\sum_{j=k}^{n}a_j\frac{j!}{(j-k)!}x^{j-k}}$$