# Evaluating the integral $\int_{0}^{\infty} \frac{\exp(-u^2)}{1+u^2} \, du$

I am trying to calculate the following integral

$$\int_{0}^{\infty} \frac{\exp(-u^2)}{1+u^2} \, du.$$

Wolfram gives a beautiful analytical answer: ${\rm e}\pi\operatorname{erfc}(1)$. I've tried every trick in my book (change of variable, contour, ...). I would love to see a proof of that beautiful result :) Thanks in advance for any help.

Define $I(s)$ by

$$I(s) = \int_{0}^{\infty} \frac{e^{-su^2}}{1+u^2} \, du.$$

Then $I(s)$ solves the following equation:

$$I(s) - I'(s) = \int_{0}^{\infty} e^{-su^2} \, du = \frac{1}{2}\sqrt{\frac{\pi}{s}}.$$

This is a 1st-order differential equation, which can be solved systematically by means of integration factor. The result is that

$$I(s) = e^s \left( \mathsf{C}-\int \frac{1}{2}\sqrt{\frac{\pi}{s}}e^{-s} \, ds \right)$$

for some appropriate choice of constant $\mathsf{C}$. Together with the boundary condition $I(\infty) = 0$, it turns out that

$$I(s) = e^s \int_{s}^{\infty} \frac{1}{2}\sqrt{\frac{\pi}{s'}}e^{-s'} \, ds' = \frac{\pi}{2}e^s \operatorname{erfc}(\sqrt{s}).$$

Plugging $s = 1$ gives the value

$$I(1) = \frac{e\pi}{2}\operatorname{erfc}(1) \approx 0.67164671082336758522\cdots.$$

• One could also use $I(0) = \tfrac\pi 2$ instead of $I(\infty) = 0$. – amsmath Sep 11 '18 at 22:16
• I was not thinking about this angle, thank you ! – None Sep 11 '18 at 22:21

By the Schwinger trick:

\begin{equation} \int_{0}^{\infty}\frac{\mathrm{e}^{-u^2}}{1+u^2}\mathrm{d}u=\int_{0}^{\infty}\mathrm{d}t\int_{0}^{\infty}\mathrm{e}^{-u^2}\mathrm{e}^{-t(1+u^2)}\mathrm{d}u= \int_{0}^{\infty}\frac{\sqrt{\pi } \mathrm{e}^{-t}}{2 \sqrt{t+1}} \mathrm{d}t=\frac{\mathrm{e\pi}}{2}\text{erfc}(1). \end{equation}

• This is close to what I had in mind, thank you ! – None Sep 20 '18 at 18:02