4
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ One could also use $I(0) = \tfrac\pi 2$ instead of $I(\infty) = 0$. $\endgroup$ – amsmath Sep 11 '18 at 22:16
  • $\begingroup$ I was not thinking about this angle, thank you ! $\endgroup$ – None Sep 11 '18 at 22:21
6
$\begingroup$

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}

$\endgroup$
  • 1
    $\begingroup$ This is close to what I had in mind, thank you ! $\endgroup$ – None Sep 20 '18 at 18:02

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.