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I'm looking to evaluate $$\int_{-\infty}^{\infty} \frac{e^{-{ax}^2}}{1+x^2} dx$$ where $a$ is real.

The "obvious" thing to try is a semi-circular contour integral in the upper or lower half plane, but this doesn't work because $e^{-{ax}^2}$ blows up along the imaginary axis.

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  • $\begingroup$ I posted an answer, but I see that this question is missing context, so I removed my answer. Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. $\endgroup$ – robjohn Dec 25 '17 at 0:54
  • $\begingroup$ @robjohn This is an integral I ran into years ago related to Kirchhoff diffraction. Wolfram Alpha gives a solution, but I was interested to see a proof. There's not much more context than that. $\endgroup$ – Yly Dec 26 '17 at 6:07
  • $\begingroup$ It would be nice to preface the question with that information. I might add some context to my answer and undelete it, as long as Mark Viola doesn't think it is too close to his answer. $\endgroup$ – robjohn Dec 26 '17 at 14:29
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Let $F(a)$ be given by the integral

$$F(a)=\int_{-\infty}^\infty \frac{e^{-ax^2}}{1+x^2}\,dx$$

for $a>0$.

Then the derivative $F'(a)$ of $F(a)$ is

$$\begin{align} F'(a)&=-\int_{-\infty}^\infty \frac{x^2e^{-ax^2}}{1+x^2}\,dx\\\\ &=F(a)-\int_{-\infty}^\infty e^{-ax^2}\,dx\\\\ &=F(a)-\sqrt{\frac{\pi}{a}} \end{align}$$

Hence, $F(a)$ satisfies the ODE $F'(a)-F(a)=-\sqrt{\frac{\pi}{a}}$ subject to $F(0)=\pi$. Solution to that ODE can be written

$$\begin{align} F(a)&=e^a \left(\pi -\sqrt \pi \int_0^a \frac{e^{-x}}{\sqrt x}\,dx\right)\\\\ &=e^a\left(\pi -\pi \frac{2}{\sqrt \pi}\int_0^{\sqrt a}e^{-x^2}\,dx\right)\\\\ &=\pi e^a \left(1-\text{erf}(\sqrt a)\right)\\\\ &=\pi e^a \text{erfc}(\sqrt a) \end{align}$$

where $\text{erf(x)}=\frac2{\sqrt \pi}\int_0^x e^{-t^2}\,dt$ is the error function and $\text{erfc}(x)=\frac2{\sqrt \pi}\int_x^\infty e^{-t^2}\,dt=1-\text{erf}(x)$ is the complementary error function.

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  • $\begingroup$ This is more of a nitpick, but I think you left out a factor of $\frac{2}{\sqrt{\pi}}$ between the first and second line of the final derivation $\endgroup$ – Dylan Dec 25 '17 at 0:01
  • $\begingroup$ (+1) I see that the answer I just posted is similar to yours, only slightly different. If it feels too close, I will delete it. $\endgroup$ – robjohn Dec 25 '17 at 0:42
  • $\begingroup$ I have removed my answer due to the fact that the question is missing context. $\endgroup$ – robjohn Dec 25 '17 at 0:55
  • $\begingroup$ Rob, thank you for the up vote. Much appreciative. $\endgroup$ – Mark Viola Dec 25 '17 at 1:29
  • $\begingroup$ @Dylan Nice catch. I've edited accordingly. $\endgroup$ – Mark Viola Dec 25 '17 at 1:29
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We need $a\geq 0$ to ensure convergence. Assuming $a>0$, the Fourier transform of $\frac{1}{1+x^2}$ is $\sqrt{\frac{\pi}{2}} e^{-|s|}$ and the Fourier transform of $e^{-ax^2}$ is $\frac{1}{\sqrt{2a}}e^{-\frac{s^2}{4a}}$, hence the original integral equals $$\sqrt{\frac{\pi}{a}}\int_{0}^{+\infty}\exp\left(-\frac{s^2}{4a}-s\right)\,ds = \color{red}{\pi e^a\text{Erfc}\sqrt{a}} $$ which is not an elementary function and depends on the CDF of a normal random variable.

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  • $\begingroup$ Nice. To clarify for those not versed in functional analysis: This argument appeals to the unitarity of the Fourier transform. The integral in question is the $L^2$ inner product of $1/(1+x^2)$ and $\exp(-ax^2)$. Taking Fourier transforms of both functions doesn't change the inner product, by unitarity. $\endgroup$ – Yly Dec 24 '17 at 23:48
  • $\begingroup$ Also, you speak of "inverse Fourier transform" of $\exp(-ax^2)$, which should properly be "Fourier transform", though it doesn't matter in this case, as the two are the same. $\endgroup$ – Yly Dec 24 '17 at 23:49
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We can apply Feynmann's Differentiation Under the Integral Trick.

Let $$ f(a)=\int_{-\infty}^\infty\frac{e^{-ax^2}}{1+x^2}\,\mathrm{d}x $$ Then $f(a)-f'(a)=\int_{-\infty}^\infty e^{-ax^2}\,\mathrm{d}x=\sqrt{\frac\pi a}$. Multiply by $-e^{-a}$ and get $$ \left(e^{-a}f(a)\right)'=-e^{-a}\sqrt{\frac\pi a} $$ Thus, $$\newcommand{\erf}{\operatorname{erf}} \begin{align} e^{-a}f(a) &=\pi-\int_0^ae^{-t}\sqrt{\frac\pi t}\,\mathrm{d}t\\ &=\pi-2\sqrt\pi\int_0^{\sqrt{a}}e^{-t^2}\,\mathrm{d}t\\[6pt] &=\pi-\pi\erf\left(\sqrt{a}\right) \end{align} $$ So that $$ f(a)=\pi e^a\left(1-\erf\left(\sqrt{a}\right)\right) $$

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