TITLE: Integral $\int_0^{\infty} te^{-\frac{1}{t^2}}\, dt$
QUESTION [0 upvotes]: I want to compute $\lim_{x\to \infty}\int_0^{x}te^{-\frac{1}{t^2}}\, dt=\int_0^{\infty} te^{-\frac{1}{t^2}}\, dt$. To do this I have thought that since I have an illimitate interval and in $0$ the function $te^{-\frac{1}{t^2}}$ is not definite I can write:
$$\int_0^{\infty} te^{-\frac{1}{t^2}}\, dt=\int_0^{1} te^{-\frac{1}{t^2}}\, dt+\int_1^{\infty} te^{-\frac{1}{t^2}}\, dt$$
The second integral is divergent since $te^{-\frac{1}{t^2}}\sim t(1-\frac{1}{t^2})=t-\frac{1}{t}$. So:
$$\int_1^{\infty} te^{-\frac{1}{t^2}}\, dt=\int_1^{\infty} t\, dt-\int_1^{\infty} \frac{1}{t}\, dt$$
and since the secon integral is divergent I have that $\int_1^{\infty} te^{-\frac{1}{t^2}}\, dt$ is divergent. So the overall integral is divergent whatever is the behaviour of $\int_0^{1} te^{-\frac{1}{t^2}}\, dt$.
But really I have to understand its behaviour:
the problem is in $0$ where the function is not defined but really the limit as $t$ goes to $0$ is finite, so how can I treat this integral?

REPLY [0 votes]: $$I=\int te^{-\frac{1}{t^2}}\, dt$$ Make one integration by parts with $u=e^{-\frac{1}{t^2}}$ and you have
$$I=\frac{1}{2} t^2\, e^{-\frac{1}{t^2}} - \int \frac{e^{-\frac{1}{t^2}}}{t}\,dt$$ Now, let $t^2=\frac{1}{y}$
$$\int \frac{e^{-\frac{1}{t^2}}}{t}\,dt=-\frac 12 \int \frac{e^{-y}}{ y}\,dy=-\frac{1}{2}\text{Ei}(-y)$$ where appears the exponential integral function.
So, back to $t$,
$$I=\int te^{-\frac{1}{t^2}}\, dt=\frac{1}{2} \left(\text{Ei}\left(-\frac{1}{t^2}\right)+e^{-\frac{1}{t^2}} t^2\right)$$
$$J=\int_0^x te^{-\frac{1}{t^2}}\, dt=\frac{1}{2} \left(\text{Ei}\left(-\frac{1}{x^2}\right)+e^{-\frac{1}{x^2}} x^2\right)$$ and the asymptotics  (already given by @heropup) is
$$J=\frac {\gamma-1}2 +\frac{x^2}{2}-\log (x)-\frac{1}{4 x^2}+\frac{1}{24 x^4}+O\left(\frac{1}{x^6}\right)$$