Integral of $x^2e^{-ax^2}$

Question

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!

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

Answer 2

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.