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I was wondering if someone could help me with the following integral:

$$ \int_{0}^{\infty}\frac{\cos(ax)}{1+x^{2n}}dx\:\:(x,\in\mathbb{C},n\in \mathbb{N},a\in \mathbb{R}) $$

I’ve tried to solve it using the 'pizza contour,' taking the real part of $e^{iax}$:

$$ I=\int_{0}^{\infty}\frac{\cos(ax)}{1+x^{2n}}dx=Re\int_{0}^{\infty}\frac{e^{iax}}{1+x^{2n}}dx=2\pi i \sum Res(\omega,z) $$

where $\omega=\frac{e^{iax}}{1+x^{2n}}dx$

Enclosing one of the poles the function has in the upper half-plane, so that there are three integrals:

$$ \int_{0}^{r}\frac{e^{iax}}{1+x^{2n}}dx\:+\:\int_{\gamma_r}\frac{e^{iax}}{1+x^{2n}}dx\:+\:\int_{r}^{0}\frac{e^{iax}}{1+x^{2n}}dx $$

The first tends to I as r tends to $\infty$, meanwhile it is well-known that the second one tends to zero as r tends to $\infty$. But the problem is with the third one, we have to parametrize $x$ as $x=t\cdot e^{i\frac{\pi}{n}}$ but I am not capable of solve that integral in terms of I.

PS: Apologies for the errors in writing and tags, this is my first post :(

RobPratt
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Pablo García García
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1 Answers1

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Employ a semi-circle contour to evaluate

$$\int_{0}^\infty \frac{\cos a x}{b^2+x^2}dx =\frac\pi {2b} e^{-ab}\tag1$$ and differentiate with respect to $a$ to get $$\int_{0}^\infty \frac{x\sin ax}{b^2+x^2}dy =\frac\pi 2 e^{-ab} \tag2$$ Partially fractionalize $I_n=\int_{0}^\infty \frac{\cos ax}{1+x^{2n}}dx $, along with $\theta_k=\frac{(2k-1)\pi}{2n}$ \begin{align} I_n=& \int_{-\infty}^\infty \frac{\cos ax}{2n}\sum_{k=1}^n \frac{1+x \cos \theta_k}{1+2x \cos \theta_k+x^2}dx\>\>\>\>\>\>\>\>x\to x-\cos \theta_k\\ =& \ \sum_{k=1}^n \int_{-\infty}^\infty \frac{\cos ax \cos(a\cos \theta_k)\sin^2 \theta_k+ x\sin ax \sin(a\cos \theta_k)\cos \theta_k}{2n(\sin^2 \theta_k+x^2)} dx\\ =&\ \frac\pi {2n}\sum_{k=1}^n e^{-a\sin \theta_k}\left[\sin\theta_k\cos(a\cos \theta_k) + \cos\theta_k\sin(a\cos \theta_k)\right]\\ =&\ \frac\pi {2n}\sum_{k=1}^n e^{-a\sin \theta_k}\sin(\theta_k+a\cos \theta_k)\\ \end{align} where (1) and (2) are used with $b=\sin\theta_k$.

Quanto
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