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Find the value of $\int_{-\infty}^{\infty} \frac{e^{-x^2}}{1+x^2} dx$

My attempt

First, I know the existence of this integral since $$ \mid \int_{-a}^{a} \frac{e^{-x^2}}{1+x^2} dx \mid \;\; \leq\;\; \mid \int_{-a}^{a} \frac{1}{1+x^2} dx \mid$$

Next, let $R >0 $ be an arbitrary and $\gamma_1(t) = -R(1-t) + Rt , \;\; 0 \leq t \leq1$ and $\gamma_2(t)= Re^{i\pi (t-1)}, \;\; 1 \leq t \leq 2 $.

So, take $f(z) = \frac{e^{-z^2}}{1+z^2}$.

I tried to find the value or upper bound of norm of $$\int_{\gamma_2} \frac{e^{-z^2}}{1+z^2} dz$$.

However, $$ \lvert\int_{\gamma_2} \frac{e^{-z^2}}{1+z^2} dz\rvert \leq \frac{\text{max}\lvert e^{-z^2}\rvert}{R^2-1}\pi R = \frac{e^{R^2}}{R^2-1}\pi R $$ So, I can't proceed this.

May I ask you how to solve this?

hew
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  • Another approach would be to use Fourier transform theory. By Parseval's theorem,

    $$\langle \mathcal{F}f,\mathcal{F}g\rangle = \langle f,g\rangle.$$

    Here $f(x) = \exp(-x^2)$ and $g(x) = \frac{1}{1+x^2}$. The Fourier transforms of these are well-known and are, respectively, $\frac{1}{\sqrt{2}}\exp(-x^2/4)$ and $\sqrt{\frac{\pi}{2}}e^{-|x|}$. You can then use basic calculus techniques to evaluate the integral

    $$\int_{-\infty}^{\infty} e^{-x^2/4} e^{-|x|}.$$

     – Cameron Williams Nov 21 '19 at 13:27
  • @CameronWilliams Thank you ! – hew Nov 26 '19 at 06:17
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    Duplicate (without complex analysis). – metamorphy Feb 27 '23 at 10:48

1 Answers1

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Tags aside, you don't explicitly say the solution needs to be by complex analysis, so I'll suggest something else. Your integral is $ef(1)$ with $f(t):=\int_{\Bbb R}\frac{e^{-t(1+x^2)}}{1+x^2}dx$ so$$f^\prime(t)=-\int_{\Bbb R}e^{-t(1+x^2)}dx=-\sqrt{\pi}t^{-1/2}e^{-t},\,f(\infty)=0.$$Hence$$f(1)=-\int_1^\infty f^\prime(t)dt=\sqrt{\pi}\int_1^\infty t^{-1/2}e^{-t}dt=2\sqrt{\pi}\int_1^\infty e^{-u^2}du=\pi\operatorname{erfc}(1).$$So the original integral is $\pi e\operatorname{erfc}(1)$.

J.G.
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