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

I found a nice formula of the following integral here

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

It states there that it can be proved by using contours method which I do not understand. It seems that the RHS is Euler's reflection formula for the gamma function but I am not so sure. Could anyone here please help me how to obtain it preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

If you don't mind, I would like to present an alternative approach that makes use of the fact that $$\int^\infty_0\frac{x^{p-1}}{1+x}dx=\frac{\pi}{\sin{p\pi}}$$ Simply factorise the denominator and decompose the integrand into partial fractions. \begin{align} \int^\infty_0\frac{x^a}{x^2+2(\cos{b})x+1}dx &=\int^\infty_0\frac{x^a}{(x+e^{ib})(x+e^{-ib})}dx\\ &=\frac{1}{-e^{ib}+e^{-ib}}\int^\infty_0\frac{x^a}{e^{ib}+x}dx+\frac{1}{-e^{-ib}+e^{ib}}\int^\infty_0\frac{x^a}{e^{-ib}+x}dx\\ &=\frac{1}{-2i\sin{b}}\int^\infty_0\frac{(e^{ib}u)^a}{1+u}du+\frac{1}{2i\sin{b}}\int^\infty_0\frac{(e^{-ib}u)^a}{1+u}du\\ &=\frac{e^{iab}}{-2i\sin{b}}\frac{\pi}{\sin(\pi a+\pi)}+\frac{e^{-iab}}{2i\sin{b}}\frac{\pi}{\sin(\pi a+\pi)}\\ &=\frac{\pi}{\sin \pi a\sin{b}}\left(\frac{e^{iab}-e^{-iab}}{2i}\right)\\ &=\frac{\pi\sin{ab}}{\sin{\pi a}\sin{b}} \end{align}

Step 1. Divide $$I=\int_0^\infty=\int_0^1+\int_1^\infty$$ and in the second integral apply the substitution $x=1/t$: $$I=\int_0^1\frac{x^\alpha \,dx}{1+2x\cos\beta+x^2}+\int_0^1\frac{t^{-\alpha} \,dt}{t^2+2t\cos\beta+1}=\int_0^1\frac{(x^\alpha+x^{-\alpha}) \,dx}{1+2x\cos\beta+x^2}.$$

Step 2. Use the following series expansion: $$\sum_{n=1}^\infty (-1)^{n+1}x^n\sin n\beta=\frac{x\sin\beta}{1+2x\cos\beta+x^2},\quad |x|<1$$ (it is easily obtained by considering imaginary part of $\sum\limits_{n=1}^\infty q^n=\frac{q}{1-q}$ with $q=-xe^{i\beta}$). Integrating term-by-term yields $$I=\frac{1}{\sin\beta}\int_0^1\sum_{n=1}^\infty (-1)^{n+1}(x^{n-1+\alpha}+x^{n-1-\alpha})\sin n\beta\,dx=\frac{1}{\sin\beta}\sum_{n=1}^\infty\frac{(-1)^{n+1}2n\sin n\beta}{n^2-\alpha^2}.$$

Step 3. Recall Fourier series expansion for the function $\sin\alpha x$: $$\sin\alpha x=\frac{\sin\pi\alpha}{\pi}\sum_{n=1}^\infty\frac{(-1)^{n+1}2n\sin nx}{n^2-\alpha^2},\quad x\in(-\pi,\pi),$$ which is proved by calculating of Fourier coefficients.

Taking here $x=\beta$ we conclude that $$I=\frac{\pi\sin\alpha\beta}{\sin\pi\alpha\sin\beta}.$$

If I may use contours to show this one.

Consider $\displaystyle I=\frac{z^{a}}{z^{2}-2z\cos(b)+1}, \;\ 0<a<1, \;\ 0<b<\pi$

Use a semicircle in the UHP with center at origin.

The poles lie at $z=e^{\pm ib}=\cos(b)\pm i\sin(b)$

Along the x axis:

$$\int_{-\infty}^{0}\frac{x^{a}}{x^{2}-2x\cos(b)+1}dx+\int_{0}^{\infty}\frac{x^{a}}{x^{2}-2x\cos(b)+1}dx$$

Let $x\to -x$ in the first integral and get:

$$\int_{0}^{\infty}\frac{x^{a}}{x^{2}-2x\cos(b)+1}dx+\int_{0}^{\infty}\frac{(-1)^{a}x^{a}}{x^{2}+2x\cos(b)+1}dx, \;\ e^{\pi ia}=(-1)^{a}$$

Round the big arc:

$$\int_{0}^{\pi}\frac{R^{a}e^{ia\theta}iRe^{i\theta}}{R^{2}e^{2i\theta}-2Re^{i\theta}\cos(b)+1}d\theta$$

which, since $a<1$, vanishes as $R\to 0$

By using $z=e^{ib}+re^{i\phi}$, the residue at $e^{ib}$ is

$$2\pi i Res(e^{ib})=2\pi i \lim_{r\to 0}\frac{(e^{ib}+re^{i\phi})^{a}}{e^{ib}+re^{i\phi}-e^{-ib}}$$

$$=\frac{2\pi i e^{iab}}{e^{ib}-e^{-ib}}=\frac{\pi}{\sin(b)}e^{iab}$$

Thus,

$$\int_{0}^{\infty}\frac{x^{a}}{x^{2}-2x\cos(b)+1}dx+e^{i\pi a}\int_{0}^{\infty}\frac{x^{a}}{x^{2}+2x\cos(b)+1}dx=\frac{\pi}{\sin(b)}e^{iab}$$

Equate real and imaginary parts:

$$\int_{0}^{\infty}\frac{x^{a}}{x^{2}-2x\cos(b)+1}dx+\cos(\pi a)\int_{0}^{\infty}\frac{x^{a}}{x^{2}+2x\cos(b)+1}dx=\frac{\pi}{\sin(b)}\cos(ab)$$

and $$\sin(\pi a)\int_{0}^{\infty}\frac{x^{a}}{x^{2}+2x\cos(b)+1}dx=\frac{\pi}{\sin(b)}\sin(ab)$$

Hence, we have a two for one:

$$\int_{0}^{\infty}\frac{x^{a}}{x^{2}+2x\cos(b)+1}dx=\frac{\pi}{\sin(b)}\frac{\sin(ab)}{\sin(\pi a)}$$ and

$$\int_{0}^{\infty}\frac{x^{a}}{x^{2}-2x\cos(b)+1}dx=\frac{\pi}{\sin(b)}\frac{\sin(a(\pi -b))}{\sin(\pi a)}$$

The last one follows from the first from writing $\pi -b$ for $b$.

The shortest method to calculate this integral is using Ramanujan's Master theorem. Write $$ \frac{1}{1+2 x \cos{(\pi \beta)} + x^2}=\sum_{n=0}^\infty \lambda(n)(-x)^n,\quad \text{near}~x=0 ~~\text{with}~~ \lambda(n)=\frac{\sin\pi\beta(n+1)}{\sin\pi\beta}. $$ Then $$ \int_0^\infty \frac{x^{\alpha}dx}{1+2x\cos\pi\beta +x^{2}}=\frac{\pi\lambda(-\alpha-1)}{\sin(\pi\alpha+\pi)}=\frac{\pi\sin\pi\alpha\beta}{\sin\pi\alpha\sin\pi\beta}. $$

Notic it(Fourier expansion):$$\frac{1}{1-2q\cos x+q^2}=\sum_{n=0}^\infty\frac{\sin nx}{\sin x} q^{n-1}$$ so:$$\int_0^\infty\frac{x^{\alpha}dx}{1+2x\cos\pi\beta+x^2} =\int_0^\infty x^{\alpha}\Big(\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}(-1)^{n-1}x^{n-1}\Big)dx$$ $$=\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}(-1)^{n-1}\int_0^\infty x^{\alpha+n-1}dx=\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}\frac{(-1)^{n-1}}{\alpha+n}=\frac{1}{\sin\pi\beta}\sum_{n=0}^\infty\frac{(-1)^{n-1}\sin n\beta}{\alpha+\beta}=\frac{\pi\sin\pi\alpha\beta}{\sin\pi\alpha\sin\pi\beta}$$