# how could we compute this infinite real integral using complex methods?

$\int^{\infty}_{-\infty} \frac{cos(x)}{x^4+1}dx$

I know a similar result, but I’m not sure if I can take it for granted, that $\int^{\infty}_{-\infty} \frac{cos(x)}{x^2+1}dx = \frac{\pi}{e}$

The section in the book is related to Cauchy’s Integral Formulas and Liuville’s th’m, but I’m not sure how to apply these here.

#### Solutions Collecting From Web of "how could we compute this infinite real integral using complex methods?"

Use contour $C$ a semicircle of radius $R$ with its bottom the $\Re z$-axis. The contour is depicted in the figure below for $R=5$.

Then
$$\oint_C\frac{e^{iz}}{z^4+1}dz=\int_{-R}^R\frac{e^{ix}}{x^4+1}dx+\int_0^{\pi}\frac{e^{-R\sin\theta+iR\cos\theta}}{R^4e^{4i\theta}+1}iRe^{i\theta}d\theta$$
Now
$$\lim_{R\rightarrow\infty}\int_{-R}^R\frac{e^{ix}}{x^4+1}dx=\int_{-\infty}^{\infty}\frac{\cos x}{x^4+1}dx$$
Because $\frac{i\sin x}{x^4+1}$ is an odd function of $x$. Also, for $R>1$,
$$\left|\int_0^{\pi}\frac{e^{-R\sin\theta+iR\cos\theta}}{R^4e^{4i\theta}+1}iRe^{i\theta}d\theta\right|\le\int_0^{\pi}\frac{R\,d\theta}{R^4-1}=\frac{\pi R}{R^4-1}\rightarrow0$$
as $R\rightarrow\infty$. So that means that
$$\int_{-\infty}^{\infty}\frac{\cos x}{x^4+1}dx=2\pi i\times\sum(\text{residues in upper half-plane})$$
There are only $2$ poles there, at $\omega_1=\frac{1+i}{\sqrt2}$ and at $\omega_2=\frac{-1+i}{\sqrt2}$. Using L’Hopital’s rule,
\begin{align}\int_{-\infty}^{\infty}\frac{\cos x}{x^4+1}dx&=2\pi i\left[\lim_{z\rightarrow\omega_1}\frac{(z-\omega_1)e^{iz}}{z^4+1}+\lim_{z\rightarrow\omega_1}\frac{(z-\omega_2)e^{iz}}{z^4+1}\right]\\ &=2\pi i\left[\lim_{z\rightarrow\omega_1}\frac{e^{iz}}{4z^3}+\lim_{z\rightarrow\omega_2}\frac{e^{iz}}{4z^3}\right]\\ &=2\pi i\left[\frac{e^{i\omega_1}}{4\omega_1^3}+\frac{e^{i\omega_2}}{4\omega_2^3}\right]=-\frac{\pi i}2\left[\omega_1e^{i\omega_1}+\omega_2e^{i\omega_2}\right]\\ &=-\frac{\pi i}2\left[\frac{1+i}{\sqrt2}e^{\frac{-1+i}{\sqrt2}}+\frac{-1+i}{\sqrt2}e^{\frac{-1-i}{\sqrt2}}\right]\\ &=-\frac{\pi i}{2\sqrt2}e^{\frac{-1}{\sqrt2}}\left[\cos\frac1{\sqrt2}+i\sin\frac1{\sqrt2}+i\cos\frac1{\sqrt2}-\sin\frac1{\sqrt2}\right]\\ &-\frac{\pi i}{2\sqrt2}e^{\frac{-1}{\sqrt2}}\left[-\cos\frac1{\sqrt2}+i\sin\frac1{\sqrt2}+i\cos\frac1{\sqrt2}+\sin\frac1{\sqrt2}\right]\\ &=\frac{\pi}{\sqrt2}e^{\frac{-1}{\sqrt2}}\left[\sin\frac1{\sqrt2}+\cos\frac1{\sqrt2}\right]\end{align}

Remark: Wolfram|Alpha agrees with this result.