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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.

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2 Answers 2

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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

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

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  • $\begingroup$ One could also use $I(0) = \tfrac\pi 2$ instead of $I(\infty) = 0$. $\endgroup$
    – amsmath
    Commented Sep 11, 2018 at 22:16
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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}

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