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.
 A: 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 $b$. 
A: 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 
$$
\int_0^\infty \frac{x^{\alpha}dx}{1+2x\cos\pi\beta +x^{2}}=\frac{\pi\lambda(-\alpha-1)}{\sin(\pi\alpha+\pi)}=\frac{\pi\sin\pi\alpha\beta}{\sin\pi\alpha\sin\pi\beta}.
$$
A: 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}
A: 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:$$\int_0^\infty\frac{x^{\alpha}dx}{1+2x\cos\pi\beta+x^2}
=\int_0^\infty x^{\alpha}\Big(\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}(-1)^{n-1}x^{n-1}\Big)dx$$
$$=\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}(-1)^{n-1}\int_0^\infty x^{\alpha+n-1}dx=\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}\frac{(-1)^{n-1}}{\alpha+n}=\frac{1}{\sin\pi\beta}\sum_{n=0}^\infty\frac{(-1)^{n-1}\sin n\beta}{\alpha+\beta}=\frac{\pi\sin\pi\alpha\beta}{\sin\pi\alpha\sin\pi\beta}$$
A: 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 $$I=\frac{\pi\sin\alpha\beta}{\sin\pi\alpha\sin\beta}.$$
A: Decomposing the integral into two, we have
$$
I= \int_0^{\infty} \frac{x^\alpha}{1+2 x \cos \beta+x^2} d x=\frac{1}{e^{\beta i}-e^{-\beta i}} \int_0^{\infty}\left(\frac{x^\alpha}{x+e^{-\beta i}}-\frac{x^\alpha}{x+e^{\beta i}}\right) d x
$$
Putting $u=\frac{x}{e^{\beta i}}$ transforms the second integral into
\begin{aligned} \int_0^{\infty} \frac{x^x}{x+e^{\beta i}} d x & =\int_0^{\infty} \frac{\left(e^{\beta i} u\right)^\alpha}{u+1} d u \\ & =e^{\alpha \beta i} \int_0^{\infty} \frac{u^\alpha}{u+1} d u \\ & =e^{\alpha \beta i} B(\alpha+1,-\alpha) \\ & =-\frac{\pi e^{\alpha \beta i}}{\sin (\pi \alpha)}\end{aligned}
Similarly, we get the first one by replacing $\beta$ by $-\beta$.
$$\int_0^{\infty} \frac{x^\alpha}{x+e^{-\beta i}} d x = -\frac{\pi e^{-\alpha \beta i}}{\sin (\pi \alpha)}$$
Plugging them together yields
$$
\boxed{I=\frac{1}{2 i \sin \beta} \frac{\pi}{\sin (\alpha \pi)}\left(e^{\alpha \beta i}-e^{-\alpha \beta i}\right)=\frac{\pi \sin (\alpha \beta)}{\sin (\alpha \pi) \sin \beta}}$$
