# 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.

## 6 Answers

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}

• +1) This derivation is great example of how a tiny bit of knowledge of complex number algebra can go a very long way. Commented Aug 8, 2014 at 5:38
• Brilliant! My feeling is right that the RHS is Euler's reflection formula for the gamma function. Thank you so much... (✿◠‿◠) Commented Aug 8, 2014 at 5:44
• How to evaluate the first integral? Commented Nov 21, 2019 at 9:40
• @MaJoad See here Commented Feb 16, 2022 at 14:50

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

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^{\alpha}}{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 $$\boxed{I=\frac{1}{2 i \sin \beta} \frac{\pi}{\sin (\alpha \pi)}\left(e^{\alpha \beta i}-e^{-\alpha \beta i}\right)=\frac{\pi \sin (\alpha \beta)}{\sin (\alpha \pi) \sin \beta}}$$

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:$$\int_0^\infty\frac{x^{\alpha}dx}{1+2x\cos\pi\beta+x^2} =\int_0^\infty x^{\alpha}\Big(\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}(-1)^{n-1}x^{n-1}\Big)dx$$ $$=\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}(-1)^{n-1}\int_0^\infty x^{\alpha+n-1}dx=\sum_{n=0}^\infty\frac{\sin n\pi\beta}{\sin\pi\beta}\frac{(-1)^{n-1}}{\alpha+n}=\frac{1}{\sin\pi\beta}\sum_{n=0}^\infty\frac{(-1)^{n-1}\sin n\beta}{\alpha+\beta}=\frac{\pi\sin\pi\alpha\beta}{\sin\pi\alpha\sin\pi\beta}$$