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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ hint: try differentation under the integral sign $\endgroup$ – tired Nov 19 '17 at 3:15
  • $\begingroup$ Look at $F(a) = \int_{-\infty}^\infty \frac{e^{-a x^2}}{1+x^2}dx, G(a) = \pi e^{-a}+F(a) = \int_{-\infty}^\infty \frac{e^{-a(1+ x^2)}}{1+x^2}dx$ and $(G(a^2))'$ $\endgroup$ – reuns Nov 19 '17 at 3:17
  • 1
    $\begingroup$ 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

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 $$F(a)=e^{a^2}\left(C-2\sqrt{\pi}\mathrm{sign}(a)\int{e^{-a^2}da}\right)$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Let $f(a)=\int_{-\infty}^\infty \frac{e^{-a^2x^2}}{1+x^2}\,dx$. Then, we have

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.