Integral asymatotics and obtaining the asym series Hi guys I am working on this problem where I am interested in finding the asymptotic series.
$I(x)= \int _0 ^x e^{s^2}s^2 ds$ as $x\rightarrow \infty$
My approach was to introduce a substatution $s =\frac{1}{u}$ that will imply that $ds = \frac{-1}{u^2}du$ thus wehen we write our integral we get
$$\int _\infty ^p e^{\frac{-1}{u^2}}du$$ 
Such taht $p \rightarrow 0$. Now I am thinking that I can just take the $e^{\frac{-1}{u^2}}$ and expand it around zero at do term by term integration. Can someone comment if this sounds good?
 A: I think that it could be easier to integrate by parts first
$$u'=e^{s^2}s\,ds \implies u=\frac{1}{2}e^{s^2}$$ $$v=s\implies v'=ds$$ which make $$\int e^{s^2}s^2\,ds=\frac{1}{2}e^{s^2} s-\frac{1}{4} \sqrt{\pi } \text{erfi}(s)$$ making $$I(x)=\frac{1}{2}e^{x^2} x-\frac{1}{4} \sqrt{\pi } \text{erfi}(x)=\frac{1}{2} e^{x^2} (x-F(x))$$ where appear the imaginary error function and the Dawson function and look at the asymptotics of these functions.
For infinitely large values of $x$, this would give $$I(x)\approx \frac 14 e^{x^2}\left(2x-\frac 1x-\frac 1{2x^3}+\cdots  \right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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You can still play with the original integral.

\begin{align}
\mrm{J}_{n}\pars{x} & \equiv \int_{0}^{x}\expo{s^{2}}s^{n}\,\dd s =
{1 \over n + 1}\int_{s = 0}^{s = x}\expo{s^{2}}\,\dd\pars{s^{n + 1}}
\\[5mm] & =
{1 \over n + 1}\,\expo{x^{2}}x^{n + 1} - {1 \over n + 1}\int_{0}^{x}s^{n + 1}
\bracks{\expo{s^{2}}\pars{2s}}\,\dd s
\\[5mm] & =
{1 \over n + 1}\,\expo{x^{2}}x^{n + 1} -
{2 \over n + 1}\,\mrm{J}_{n + 2}\pars{x}
\\[5mm] \implies
\mrm{J}_{n}\pars{x} & = {1 \over 2}\,\expo{x^{2}}x^{n - 1} -
{1 \over 2}\,\pars{n - 1}\,\mrm{J}_{n - 2}\pars{x}\,,\qquad
\left\{\begin{array}{rcl}
\ds{\mrm{J}_{0}\pars{x}} & \ds{=} & \ds{{1 \over 2}\,\root{\pi}\,\mrm{erfi}\pars{x}}
\\[2mm]
\ds{\mrm{J}_{1}\pars{x}} & \ds{=} & \ds{{1 \over 2}\,\pars{\expo{x^{2}} - 1}} 
\end{array}\right.
\end{align}
