integrate $\int_{-\infty}^\infty \frac{e^{\frac{-1}{1+x^2}}}{1+x^2}dx$ using complex integration

Question

I would like to calculate the integral

$$ \int_{-\infty}^\infty \frac{e^{\frac{-1}{1+x^2}}}{1+x^2}dx. $$

Using contour integration and the residue theorem I managed to show that it is equal to

$$2\pi i \, Res(\frac{e^{\frac{-1}{1+z^2}}}{1+z^2},i) $$ However, as this is not a pole, and I see no apparent decomposition to known Laurent series, I'm not sure how to calculate the residue.

Answer 1

Let's continue your evaluation. We have $$\int_{-\infty}^{\infty}\frac{e^{-1/(1+x^2)}}{1+x^2}\,dx=\oint_{\gamma}\frac{e^{-1/(1+z^2)}}{1+z^2}\,dz$$ where a simple closed contour $\gamma$ encircles $z=i$ (but not $z=-i$).

Substituting $z=i\frac{1+w}{1-w}$ (i.e. $w=\frac{z-i}{z+i}$), we get $$\int_{-\infty}^{\infty}\frac{e^{-1/(1+x^2)}}{1+x^2}\,dx=\frac{e^{-1/2}}{2i}\oint_{\gamma'}e^{(w+w^{-1})/4}\frac{dw}{w}$$ with $\gamma'$ encircling $w=0$. Now the Laurent series we need is $$\exp\left[\frac{z}{2}\Big(w+\frac{1}{w}\Big)\right]=\sum_{n\in\mathbb{Z}}I_n(z)w^n.$$ So the integral is $\pi e^{-1/2}I_0(1/2)$.

Answer 2

With substitution $x=\tan(t/2)$ $$\int_{-\infty}^\infty \frac{e^{\frac{-1}{1+x^2}}}{1+x^2}dx=\dfrac{1}{2\sqrt{e}}\int_{-\pi}^{\pi}e^{-\frac12\cos t}\ dt= \dfrac{\pi}{\sqrt{e}}I_0(\frac12)$$