$\int_{-\infty}^\infty \frac{e^{-\lvert x \rvert}}{x^4+1}dx$

Question

$$\int_{-\infty}^\infty \frac{e^{-\lvert x \rvert}}{x^4+1}dx$$

Wolframalpha says the answer is 1.26096, which is not helpful.

I tried with contour integral, but I should divide $x > 0$ and $x < 0$ domain and I'm not sure how to proceed from there, I mean, don't know how to draw the contour. I don't think I can draw contour through imaginary axis cuz the integral would diverge there.

Answer 1

I know not of nice closed form, but one has

$$\frac1{x^4+1}=\frac14\sum_{\omega^4=-1}\frac\omega{x+\omega}$$

and

$$\int_0^\infty\frac{e^{-x}}{x+\omega}~dx=e^\omega E_1(\omega)$$

where $E_1$ is an exponential integral. Hence, your integral is given by

$$\int_{-\infty}^\infty\frac{e^{-|x|}}{x^4+1}~dx=\frac12\sum_{\omega^4=-1}\omega e^\omega E_1(\omega)$$

Answer 2

Just an addendum to SBA's answer.

$$ \int_{\mathbb{R}}\frac{e^{-|x|}}{x^4+1}\,dx = 2\sqrt{2}\int_{\mathbb{R}}\frac{e^{-|x|/\sqrt{2}}}{(x^2+2x+2)(x^2-2x+2)}\,dx $$ boils down to $$ \frac{1}{2\sqrt{2}}\int_{\mathbb{R}}e^{-|x|/\sqrt{2}}\left(\frac{2-x}{x^2-2x+2}+\frac{2+x}{x^2+2x+2}\right)\,dx $$ then to (by parity) $$ \frac{1}{\sqrt{2}}\int_{0}^{+\infty}e^{-x/\sqrt{2}}\left(\frac{2-x}{x^2-2x+2}+\frac{2+x}{x^2+2x+2}\right)\,dx. $$ Through the Fourier transform the original integral can be written as $$ I=2\int_{0}^{+\infty}\frac{\cos(s)+\sin(s)}{e^s(1+2s^2)}\,ds $$ and easily approximated through the Cauchy-Schwarz inequality ($I^2\leq\frac{3\pi}{4\sqrt{2}}$), since the new integrand function has a Gaussian behaviour in a neighbourhood of the origin, where most of the mass of the integral is concentrated. A not terrible approximation is provided by $$ I \approx 2\int_{0}^{+\infty}\frac{ds}{\left(1+\frac{3}{2}s^2\right)^2} = \frac{\pi}{\sqrt{6}}. $$ I am not expecting a closed form beside the one provided by the exponential integral.