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$$ \int_{-\infty}^{\infty} \frac{x^2e^{-\alpha x^2}}{x^2+b^2} dx $$

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    contour integration? – 2017-01-11
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    Please spare us you quick drafts. – 2017-01-11
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    We can transform the problem into solving an ODE $$f'(a) - b^2 f(a) + \frac{\sqrt{\pi}}{2a^{3/2}} = 0$$ by defining the integral as $f(a)$ and using $f(a) = -\frac{d}{da}\int \frac{e^{-ax^2}}{x^2+b^2} = -\frac{1}{b^2}\frac{d}{da}\left[\int e^{-ax^2} - f(a)\right]$ together with $\int e^{-ax^2} = \sqrt{\pi/a}$. – 2017-01-11
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    this integral containes the error function – 2017-01-11
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    @YvesDaoust He is required to show his work, and he is doing so. – 2017-01-11
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    @GFauxPas: we spend lots of time laying out our solutions properly. – 2017-01-11
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    Thank you @Winther, I would never think to transform this into an ODE. – 2017-01-11

2 Answers 2

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A 'sort-of' elementary way to proceed [without complex analysis] is as follows. Write $$ \frac{1}{x^2+b^2}=\int_0^\infty ds\ e^{-s(x^2+b^2)} $$ and exchange the order of integrals to get $$ I= \int_0^\infty ds\ e^{-s b^2}\int_{-\infty}^\infty dx\ x^2 e^{-(\alpha+s)x^2} $$ $$ =\partial_\alpha\int_0^\infty ds\ e^{-s b^2}\int_{-\infty}^\infty dx\ e^{-(\alpha+s)x^2} $$ $$ =\partial_\alpha\int_0^\infty ds\ e^{-s b^2}\sqrt{\frac{\pi}{\alpha+s}} $$ and changing variables $\alpha+s=u$ $$ =\sqrt{\pi}\partial_\alpha \left[e^{\alpha b^2}\int_\alpha^\infty du\frac{e^{-b^2 u}}{\sqrt{u}}\right]\ . $$ The integrand admits a quite simple antiderivative $$ \int du \frac{e^{-b^2 u}}{\sqrt{u}}=\frac{\sqrt{\pi }\ \text{erf}\left(b \sqrt{u}\right)}{b}+C\ , $$ in terms of the error function. The finishing touch should be easy.

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Assuming $a,b\in\mathbb{R}^+$ the given integral is simply $$\begin{eqnarray*} 2\int_{0}^{+\infty}\frac{x^2}{x^2+b^2}e^{-a x^2}\,dx &\stackrel{x\mapsto bz}{=}&2b\int_{0}^{+\infty}\frac{z^2}{z^2+1}e^{-a b^2 z^2}\,dz\\&=&\sqrt{\frac{\pi}{a}}-b\int_{-\infty}^{+\infty}\frac{\exp\left(-a b^2 z^2\right)}{z^2+1}\,dz\end{eqnarray*}$$ and we may tackle the last integral through the Fourier transform: $$ \int_{-\infty}^{+\infty}\frac{\exp\left(-a b^2 z^2\right)}{z^2+1}\,dz =\sqrt{\frac{\pi}{ab^2}} \int_{0}^{+\infty}e^{-s}\exp\left(-\frac{s^2}{4ab^2}\right)\,ds$$ getting a value of the (complementary) error function by completing the square: $$\boxed{\int_{-\infty}^{+\infty}\frac{x^2}{x^2+b^2}e^{-a x^2}\,dx = \color{red}{\sqrt{\frac{\pi}{a}}-\pi b e^{ab^2}\text{Erfc}\left(b\sqrt{a}\right)}.}$$