INTEGRAL \int_0^\infty \frac{x^{a}}{x^2+2x \cos(\theta)+1}\,dx

In this post we will evaluate the integral

$\int_0^\infty \frac{x^{\alpha}}{x^2+2x \cos(\pi \beta)+1}\,dx=\frac{\pi}{\sin\left(\pi \beta \right)} \frac{\sin( \pi \alpha \beta)}{\sin\left(\alpha\pi\right)}$

Consider the following integral

$\begin{aligned} I&=\int_0^\infty \frac{x^{a}}{x^2+2x \cos(\theta)+1}\,dx\\ &=\int_0^1 \frac{x^{a}}{x^2+2x \cos(\theta)+1}\,dx+\int_1^\infty \frac{x^{a}}{x^2+2x \cos(\theta)+1}\,dx\\ &=\int_0^1 \frac{x^{a}}{x^2+2x \cos(\theta)+1}\,dx+\int_0^1 \frac{x^{-a}}{x^2+2x \cos(\theta)+1}\,dx\\ &=\int_0^1 \frac{x^{a}+x^{-a}}{x^2+2x \cos(\theta)+1}\,dx\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{n=1}^\infty(-1)^{n-1} \sin(n \theta)\int_0^1 \left(x^{n+a-1}+x^{n-a-1} \right)\,dx\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{n=1}^\infty(-1)^{n-1} \sin(n \theta) \left(\frac{1}{n+a}+\frac{1}{n-a} \right)\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{n=1}^\infty(-1)^{n-1} \sin(n \theta) \left(\frac{2n}{n^2-a^2} \right)\\ &=\frac{2}{\sin\left(\theta \right)} \sum_{n=1}^\infty \frac{(-1)^{n-1} n\sin(n \theta)}{n^2-a^2} \qquad(\text{fourier series of } \sin(a \theta))\\ &=\frac{\pi}{\sin\left(\theta \right)} \frac{\sin(a\theta)}{\sin\left(a\pi\right)} \qquad \blacksquare \end{aligned}$

Setting $\theta=\pi \beta$ we obtain the desired result

$\boxed{\int_0^\infty \frac{x^{\alpha}}{x^2+2x \cos(\pi \beta)+1}\,dx=\frac{\pi}{\sin\left(\pi \beta \right)} \frac{\sin( \pi \alpha \beta)}{\sin\left(\alpha\pi\right)}}$

Appendix

Fourier Series $\sin(ax)$ in $[-\pi,\pi ]$

A Fourier series representation is given by the following expression:

$f(x) = \frac{a_0}{2} + \sum _ {k=1}^{\infty}\left(a_k\cos(kx) + b_k\sin (kx)\right)$

where

$a_0 = \frac{1}{\pi} \int^{\pi}_{-\pi} f(x)dx$

$a_k = \frac{1}{\pi} \int^{\pi}_{-\pi} f(x)\cos (kx)dx$

$b_k = \frac{1}{\pi} \int^{\pi}_{-\pi} f(x)\sin (kx)dx$

Lets choose $f(x)=\sin(ax)$ , then

$\begin{aligned} a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(a x)\,dx\\ &=\frac{1}{\pi}\int_{-\pi}^{0}\sin(a x)\,dx+\frac{1}{\pi}\int_{0}^{\pi}\sin(a x)\,dx\\ &=-\frac{1}{\pi}\int_{0}^{\pi}\sin(a x)\,dx+\frac{1}{\pi}\int_{0}^{\pi}\sin(a x)\,dx\\ &=0 \end{aligned}$

$\begin{aligned} a_k&=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(a x)\cos(kx)\,dx\\ &=\frac{1}{\pi}\int_{-\pi}^{0}\sin(a x)\cos(kx)\,dx+\frac{1}{\pi}\int_{0}^{\pi}\sin(a x)\cos(kx)\,dx\\ &=-\frac{1}{\pi}\int_{0}^{\pi}\sin(a x)\cos(kx)\,dx+\frac{1}{\pi}\int_{0}^{\pi}\sin(a x)\cos(kx)\,dx\\ &=0 \end{aligned}$

$\begin{aligned} b_k&=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(a x)\sin(kx)\,dx\\ &=\frac{1}{2\pi}\int_{-\pi}^{\pi}\cos\left((a-k)x\right)\,dx-\frac{1}{2\pi}\int_{-\pi}^{\pi}\cos\left((a+k)x\right)\,dx\\ &=\frac{1}{2(a-k)\pi}\int_{-\pi(a-k)}^{\pi(a-k)}\cos(x)\,dx-\frac{1}{2(a+k)\pi}\int_{-\pi(a+k)}^{\pi(a+k)}\cos(x)\,dx\\ &=\frac{1}{2(a-k)\pi}\left(\sin(x)\Big|_{-\pi(a-k)}^{\pi(a-k)}\right)-\frac{1}{2(a+k) \pi}\left(\sin(x)\Big|_{-\pi(a+k)}^{\pi(a+k)}\right)\\ &=\frac{1}{2(a-k)\pi}\left(\sin\left(\pi(a-k)\right)-\sin\left(-\pi(a-k)\right)\right)-\frac{1}{2(a+k)\pi}\left(\sin\left(\pi(a+k)\right)-\sin\left(-\pi(a+k)\right)\right)\\ &=\frac{\sin\left(\pi(a-k)\right)}{(a-k)\pi}-\frac{\sin\left(\pi(a+k)\right)}{(a+k)\pi}\\ &=\frac{\sin\left(a\pi\right)\cos(\pi k)}{(a-k)\pi}-\frac{\sin\left(a\pi\right)\cos(\pi k)}{(a+k)\pi}\\ &=\frac{(-1)^k\sin\left(a\pi\right)}{(a-k)\pi}-\frac{(-1)^k\sin\left(a\pi\right)}{(a+k)\pi}\\ &=\frac{2}{\pi} \cdot\frac{(-1)^k k\sin\left(a\pi\right)}{a^2-k^2} \end{aligned}$

Therefore

$\sin(ax)=\frac{2\sin\left(a\pi\right)}{\pi} \sum_{k=1}^\infty \frac{ (-1)^{k-1} k \sin\left(k\pi\right) }{k^2-a^2}$

$\frac{\pi\sin(ax)}{2\sin\left(a\pi\right)}= \sum_{k=1}^\infty \frac{ (-1)^{k-1} k \sin\left(k\pi\right) }{k^2-a^2}$

Evaluating the product

$\begin{aligned} P&= \left(\sum_{n=0}^\infty (-1)^ne^{-i \theta n}x^n\right) \cdot \left(\sum_{n=0}^\infty (-1)^ne^{i \theta n}x^n\right)\\ &=\left(\sum_{n=0}^\infty (-e^{-i \theta })^nx^n\right) \cdot \left(\sum_{n=0}^\infty (-e^{i \theta })^nx^n\right) \\ \end{aligned}$

Taking the Cauchy product, We have that $a_k=(-e^{-i \theta })^k$ and $b_k=(-e^{i \theta })^k$ , we therefore obtain

$\begin{aligned} P&= \left(\sum_{n=0}^\infty (-1)^ne^{-i \theta n}x^n\right) \cdot \left(\sum_{n=0}^\infty (-1)^ne^{i \theta n}x^n\right)\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n (-e^{-i \theta })^k(-e^{i \theta })^{n-k} \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n\left(\sum_{k=0}^n e^{i n \theta }e^{-2i k \theta } \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\left(\sum_{k=0}^n e^{-2i k \theta } \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\left(\frac{1-e^{-2i (n+1) \theta }}{1-e^{-2i \theta }} \right)x^n \qquad \left(\text{geometric sum} \right)\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\frac{e^{i \theta}}{e^{i \theta}}\left(\frac{1-e^{-2i (n+1) \theta }}{1-e^{-2i \theta }} \right)x^n \\ &=\sum_{n=0}^\infty(-1)^n\left(\frac{e^{i (n+1)\theta}-e^{-i (n+1) \theta }}{e^{i \theta}-e^{-i \theta }} \right)x^n \\ &=\sum_{n=0}^\infty(-1)^n\left(\frac{\sin\left((n+1)\theta \right)}{\sin\left(\theta \right)} \right)x^n \\ &=\sum_{n=1}^\infty(-1)^{n-1}\left(\frac{\sin\left(n\theta \right)}{\sin\left(\theta \right)} \right)x^{n-1}\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{n=1}^\infty(-1)^{n-1} \sin(n \theta) x^{n-1}\qquad \blacksquare\\ \end{aligned}$

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