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When working a proof, I reached an expression similar to this:

$$\int_{-\infty}^{\infty} \frac{\mathrm{e}^{-a^2 x^2}}{1 + x^2} \mathrm{d}x$$

I've tried the following:

1. I tried squaring and combining and converting to polar coordinates, like one would solve a standard Gaussian. However, this yielded something which seems no more amenable to a solution:

$$\int_{\theta=0}^{\theta=2\pi} \int_{0}^{\infty} \frac{r \mathrm{e}^{-a^2 r^2}}{(1 + r^2 \sin^2(\theta))(1 + r^2 \cos^2(\theta))} \mathrm{d}r \mathrm{d}\theta$$

2. I tried doing a trig substitution, t = tan u, and I have no idea what to do from there.

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm{e}^{-a^2 \tan^2(u)} \mathrm{d}u$$

3. I looked into doing $u^2 = 1 + x^2$ but this gives us a ugly dx that I don't know how to handle, and moreover, I think I'm breaking my limits of integration (because Mathematica no longer solves it.):

$$u^2 = 1 + x^2$$

$$2 u \mathrm{d}u = 2 x \mathrm{d}x$$

$$\mathrm{d}x = \frac{u}{\sqrt{u^2 - 1}}$$ $$\mathrm{e}^{a^2} \int_{-\infty}^{\infty} \frac{\mathrm{e}^{-a^2 u^2}}{u \sqrt{u^2 - 1}} \mathrm{d}u$$

4. I looked into some form of differentiation under the integral, but that didn't seem to yield anything that looked promising. (I checked parameterizing x^2 to x^b in both places, and in either place, and nothing canceled cleanly.)

I have a solution from Mathematica, it's:

$$\pi e^{a^2} \text{erfc}(a)$$

But I'd like to know how to arrive at this. I'm sure it's something simple I'm missing.

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  • $\begingroup$ have you tried pulling out a factor of pi? $\endgroup$ – The Great Duck Nov 19 '17 at 2:49
  • $\begingroup$ From where? I can pull out a factor of pi from part 3, but that just gives me an integral that I can't solve that is supposedly equal to erfc(a), but doesn't look like any form of erfc(a) that I recognize. $\endgroup$ – OmnipotentEntity Nov 19 '17 at 2:54
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    $\begingroup$ This is a good example where we can use Parseval's theorem together with : $e^{-\pi x^2}, e^{-\pi \xi^2}$ and $\frac{1/\pi}{1+x^2}, e^{-2\pi |\xi|}$ are Fourier transform pairs. Thus it reduces to $\int_0^\infty e^{-2 \pi \xi -\pi \xi^2}d\xi$ which is easily seen to be what mathematica claims. $\endgroup$ – reuns Nov 19 '17 at 2:59
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    $\begingroup$ hint: try differentation under the integral sign $\endgroup$ – tired Nov 19 '17 at 3:15
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    $\begingroup$ Also, complex integration with a contour on upper half plane gives us the result as we have $z=i$ in this domain! $\endgroup$ – Nosrati Nov 19 '17 at 3:21
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Let $F$ be the function $$F(a)=\int_{-\infty}^{\infty}\frac{e^{-a^{2}x^{2}}}{1+x^2}dx$$ We take the derivative w.r.t $a$ $$F^{\prime}(a)=\frac{d}{da}\left(\int_{-\infty}^{\infty}\frac{e^{-a^{2}x^{2}}}{1+x^2}dx\right)=\int_{-\infty}^{\infty}\frac{d}{da}\left(\frac{e^{-a^{2}x^{2}}}{1+x^2}\right)dx =\int_{-\infty}^{\infty}\frac{-2ax^{2}e^{-a^{2}x^{2}}}{1+x^2}dx$$ $$=\int_{-\infty}^{\infty}\frac{-2a\big((x^{2}+1)-1\big)e^{-a^{2}x^{2}}}{1+x^2}dx =-2a\int_{-\infty}^{\infty}e^{-a^{2}x^{2}}dx+2aF(a) =-2a\sqrt{\frac{\pi}{a^2}}+2aF(a)$$ Then $$F^{\prime}(a)=2a\left(F(a)-\sqrt{\pi}\,\frac{1}{\vert{a}\vert}\right) =2aF(a)-2\sqrt{\pi}\mathrm{sign}(a).$$ Then you have a differential equation: $$ F^{\prime}(a)-2a\,F(a)=-2\sqrt{\pi}\mathrm{sign}(a) $$ with initial condition $F(0)=\pi$. This fisrt order ode has integrant factor: $$\mu(a)=\displaystyle{e^{\displaystyle{\int{-2ada}}}}=e^{-a^2}$$ Then $$ \left(e^{-a^2}F(a)\right)^{\prime}=-2\sqrt{\pi}\mathrm{sign}(a) e^{-a^2} $$ this implies $$ e^{-a^2}F(a)=-2\sqrt{\pi}\int{\mathrm{sign}(a) e^{-a^2}}da+C $$ Finaly $$F(a)=e^{a^2}\left(C-2\sqrt{\pi}\mathrm{sign}(a)\int{e^{-a^2}da}\right)$$

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  • $\begingroup$ Excellent work. This simplifies down (for a > 0) to $\pi \mathrm{e}^{a^2} (-\mathrm{erf}(a) + C)$, but it's not at all clear to me how to obtain C (which should be 1). $\endgroup$ – OmnipotentEntity Nov 19 '17 at 4:19
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    $\begingroup$ Just, use that $F(0)=\pi$. :) $\endgroup$ – Hector Blandin Nov 19 '17 at 4:21
  • $\begingroup$ Oh of course facepalm I was substituting back too soon, and I had an extra factor of "a" in the derivation. How foolish. Thanks so much. $\endgroup$ – OmnipotentEntity Nov 19 '17 at 4:23
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Let $f(a)=\int_{-\infty}^\infty \frac{e^{-a^2x^2}}{1+x^2}\,dx$. Then, we have

$$\begin{align} f(a)&=2\int_0^\infty e^{-a^2x^2}\int_0^\infty e^{-s(1+x^2)}\,ds\,dx\\\\ &=2\int_0^\infty e^{-s}\int_0^\infty e^{-(s+a^2)x^2}\,dx\,ds\\\\ &=\int_0^\infty e^{-s} \frac{\sqrt{\pi}}{\sqrt{s+a^2}}\,ds\\\\ &=\sqrt{\pi}e^{a^2}\int_0^\infty \frac{e^{-(s+a^2)}}{\sqrt{s+a^2}}\,ds\\\\ &=\sqrt{\pi}e^{a^2}\int_{a^2}^\infty \frac{e^{-t}}{\sqrt t}\,dt\\\\ &=2\sqrt{\pi}e^{a^2}\int_{|a|}^\infty e^{-u^2}\,du\\\\ &=\pi e^{a^2}\text{erfc}(|a|) \end{align}$$

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Here is another approach where we first make use of an auxiliary function.

Before proceeding we recall the definitions for the error function $\text{erf}(x)$ and the complementary error function $\text{erfc}(x)$: $$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} \, dt$$ and $$\text{erfc}(x) = \frac{2}{\sqrt{\pi}} \int^\infty_x e^{-t^2} \, dt,$$ respectively such that $$\text{erf}(x) = 1 - \text{erfc}(x).$$

The idea here is to consider an auxiliary function, related to our function of interest $f(a)$, but which turns our to be a constant function for all $a$ in its domain.

Start by considering the following auxiliary function \begin{equation} I(a) = \left (\int^a_0 e^{-t^2} \, dt \right )^2 + \int^1_0 \frac{e^{-a^2 (t^2 + 1)}}{1 + t^2} \, dt, \,\, a > 0. \tag1 \end{equation} Note the term appearing between the brackets is nothing more than the error function. On differentiating the auxiliary function with respect to $a$ we obtain $$I'(a) = 2 e^{-a^2} \int^a_0 e^{-t^2} \, dt - 2a e^{-a^2} \int^1_0 e^{-a^2 t^2} \, dt.$$ In obtaining this result, Leibniz' rule for differentiating under the integral sign has been used. In the second integral, if a substitution of $u = at$ is made, the result $I'(a) = 0$ quickly follows showing the auxiliary function is indeed constant for all $a > 0$. To find the value for this constant, letting $a \to 0^+$ gives $$I(a) \to \int^1_0 \frac{dt}{1 + t^2} = \frac{\pi}{4},$$ so that $I(a) = \pi/4$ for all $a > 0$.

As the first of the integrals appearing in (1) can be written in terms of the error function we have \begin{equation} \int^1_0 \frac{e^{-a^2 (t^2 + 1)}}{1 + t^2} \, dt = \frac{\pi}{4} \left (1 - \text{erf}^2 (a) \right ). \tag2 \end{equation}

A similar thing can be done for the complementary error function. In this case we start by considering the following auxiliary function \begin{equation} J(a) = \left (\int^\infty_a e^{-t^2} \, dt \right )^2 - \int^\infty_1 \frac{e^{-a^2 (t^2 + 1)}}{1 + t^2} \, dt, \,\, a > 0. \tag3 \end{equation} Again observe the term appearing between the brackets in nothing more than the complementary error function. On differentiating with respect to $a$ we have $$J'(a) = -2 e^{-a^2} \int^\infty_a e^{-t^2} \, dt + 2a e^{-a^2} \int^\infty_1 e^{-a^2 t^2} \, dt.$$ A substitution of $u = at$ in the second integral once again reduces the derivative of the auxiliary function to zero, showing $J(a)$ is constant. Letting $a \to \infty$ in (3) we see $J(a) \to 0$. Thus $J(a) = 0$ for all $a > 0$. Writing the first of the integrals in (3) in terms of the complementary error function, one finds \begin{equation} \int^\infty_1 \frac{e^{-a^2 (1 + t^2)}}{1 + t^2} \, dt = \frac{\pi}{4} \text{erfc}^2 (a). \tag4 \end{equation}

Adding (2) to (4) yields \begin{align*} \int^\infty_0 \frac{e^{-a^2(1 + t^2)}}{1 + t^2} \, dt &= \frac{\pi}{4} \left [\text{erfc}^2 (a) + 1 - \text{erf}^2 (a) \right ]\\ &= \frac{\pi}{4} \left [\text{erfc}^2 (a) + 1 - (1 - \text{erfc}(a))^2 \right ]\\ &= \frac{\pi}{2} \text{erfc} (a). \end{align*} Rearranging gives $$e^{-a^2} \int^\infty_0 \frac{e^{-a^2 t^2}}{1 + t^2} \, dt = \frac{\pi}{2} \text{erfc}(a),$$ or $$\int^\infty_0 \frac{e^{-a^2 t^2}}{1 + t^2} \, dt = \frac{\pi}{2} e^{a^2} \text{erfc}(a).$$

So for $f(a)$ we finally have $$f(a) = 2 \int^\infty_0 \frac{e^{-a^2 t^2}}{1 + t^2} \, dt = \pi e^{a^2} \text{erfc}(a), \quad a > 0.$$

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