Integral of $x^2e^{-ax^2}$ Hey guys I need to find the following integral using integration by parts and not the gamma function. Also there is an a constant a in the exponential function. So it is actually $x^2e^{-ax^2}$.

Thanks for the help!
 A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}x^{2}
\expo{-ax^{2}}\,\dd x} =
-\,\partiald{}{a}\int_{-\infty}^{\infty}
\expo{-ax^{2}}\,\dd x
\\[5mm] = &\
-\,\partiald{}{a}\bracks{a^{-1/2}\
\overbrace{\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x}
^{\ds{=\ \root{\pi}}}} =
-\pars{-\,\half\,a^{-3/2}}\root{\pi}
= \color{#00f}{\root{\pi} \over 2a^{3/2}}
\end{align}
Also
\begin{align}
\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x&=
\pars{\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x
\int_{-\infty}^{\infty}\expo{-y^{2}}\,\dd y}^{1/2}
\\[5mm] & =
\pars{\int_{0}^{\infty}\expo{-r^{2}}r\,\dd r
\int_{0}^{2\pi}\dd\theta}^{1/2}
\\[3mm]&=\pars{\left.2\pi\,{\expo{-r^{2}} \over -2}
\right\vert_{\,0}^{\,\infty}}^{1/2}
=\root{\pi}
\end{align}
A: Perform an integration by parts. Your result is
$$\int_{-\infty}^\infty x^2 e^{-ax^2}dx = -{x\over 2a}e^{-ax^2}\bigg|_{-\infty}
^\infty +
{1\over 2a}\int_{-\infty}^\infty e^{-ax^2}dx
= {1\over 2a}\int_{-\infty}^\infty e^{-ax^2}dx$$
Now you are stuck with an integral not resolvable by elementary means.  Either you use the $\Gamma$ function or the old polar coordinate trick.
