Integral of the day XXX

An integral involving an orthogonal system

Today we show the proof of this integral posted by @integralsbot \[ \int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} = \frac{\pi (1+a^nb^{m})}{(1-a^2)(1-b^2)(1-a^nb^{m})} \quad \] \[ m,n\in \mathbb{Z}_{>0} \textrm{ with } \gcd(m,n) = 1 \textrm{ and } |a|,|b|\leq 1\] The proof relies on Fourier series and the theory of orthogonal functions.

Proof:

We will use the following Fourier series (Proof in the appendix): \[ 1+2\sum_{k=1}^{\infty} p^k\cos(kx) = \frac{1-p^2}{1-2p\cos x + p^2} \quad |p|\leq | \] Let $\displaystyle m,n\in \mathbb{Z}_{>0}$ with $\gcd(m,n) = 1 $ and $|a|,|b|\leq 1$ Then \begin{align*} \int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} =& \frac{1}{(1-a^2)(1-b^2)}\int_{0}^{\pi} \frac{1-a^2}{(1-2a\cos(mx)+a^2)}\frac{1-b^2}{(1-2b\cos(mx)+b^2)}dx\\ =& \frac{1}{(1-a^2)(1-b^2)}\int_{0}^{\pi} \left( 1+2\sum_{k=1}^{\infty} a^k\cos(kmx)\right)\left( 1+2\sum_{j=1}^{\infty} b^j\cos(jnx)\right) dx\\ =& \frac{1}{(1-a^2)(1-b^2)}\left[\int_{0}^{\pi}dx + 2\sum_{k=1}^{\infty}a^k\int_{0}^{\pi} \cos(kmx)dx+ 2\sum_{j=1}^{\infty}b^j\int_{0}^{\pi} \cos(jnx)dx + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx \right] \end{align*} Recall that a system of real functions \[\phi_{0}(x), \phi_{1}(x), \phi_{2}(x),..., \phi_{n}(x),...\] is said to be orthogonal on the interval $\left[a,b\right]$ if \[ \int_{a}^{b} \phi_{n}(x) \phi_{m}(x) dx = 0 \quad n\neq m, m=0,1,2,...\] We shall also assume that \[ \int_{a}^{b} \phi_{n}^{2}(x) dx\neq 0 \quad n=0,1,2,...\] Consider the system \[ 1,\cos(x), \cos(2x),...,\cos(nx),...\] This sytem is orthogonal on the interval $\left[0,\pi\right]$. In fact \[ \int_{0}^{\pi} \cos(nx) dx = \left[\frac{\sin(nx)}{n}\right]_{0}^{\pi} = 0 \quad n=1,2,... \tag{1}\] which means that the functions $\cos(nx)$ and $1$ are orthogonal. Moreover, we have \begin{align*} \int_{0}^{\pi} \cos(nx)\cos(mx)dx =& \frac{1}{2} \int_{0}^{\pi} \left[\cos(n+m)x + \cos(n-m)x\right] dx \\ =& \frac{1}{2}\int_{0}^{\pi} \cos(n+m)x dx + \frac{1}{2}\int_{0}^{\pi} \cos(n-m)x dx =0 \quad n\neq m\\ =& 0 \tag{2} \end{align*} Finally, note that \[ \int_{0}^{\pi} \cos^2(nx) dx = \frac{1}{2} \int_{0}^{\pi} 1+\cos(2nx) dx = \frac{\pi}{2} \tag{3}\] Hence, \begin{align*} \int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} =& \frac{1}{(1-a^2)(1-b^2)}\left[\int_{0}^{\pi}dx + 2\sum_{k=1}^{\infty}a^k\underbrace{\int_{0}^{\pi} \cos(kmx)dx}_{=0}+ 2\sum_{j=1}^{\infty}b^j\underbrace{\int_{0}^{\pi} \cos(jnx)dx}_{=0} + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx \right] \quad \textrm{from (1)} \\ =& \frac{1}{(1-a^2)(1-b^2)}\left[\pi + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\underbrace{\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx}_{J} \right] \end{align*} From (2) note that \[J = \int_{0}^{\pi} \cos(kmx)\cos(jnx) dx = \begin{cases} 0 \textrm{ if } & km \neq jn \\ \frac{\pi}{2} \textrm{ if }& km = jn \end{cases} \] However, since $\gcd(m,n)=1$ then $ n\nmid m \; \wedge \; m \nmid n $ Therefore \[ km = jn \Longleftrightarrow k = in \wedge j=im \quad i\in \mathbb{Z}\setminus\left\{0\right\} \] Hence if $\displaystyle k = in \wedge j=im$ with $ i\in \mathbb{Z}_{>0} \textrm{ and } |a|,|b|\lt 1 $ \begin{align*} \sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx =& \sum_{i=1}^{\infty}a^{in}b^{im}\int_{0}^{\pi} \cos(inmx)\cos(imnx) dx\\ =& \sum_{i=1}^{\infty}a^{in}b^{im}\int_{0}^{\pi} \cos^2(inmx) dx\\ =& \frac{\pi}{2} \sum_{i=1}^{\infty}a^{in}b^{im} \quad \textrm{ from (3) } \\ =& \frac{\pi}{2} \sum_{i=1}^{\infty}\left(a^{n}b^{m}\right)^{i}\\ =& \frac{\pi}{2} \frac{a^nb^{m}}{(1-a^nb^{m})} \tag{4} \end{align*} Therefore \begin{align*} I =& \frac{1}{(1-a^2)(1-b^2)}\left[\pi + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx \right]\\ =& \frac{1}{(1-a^2)(1-b^2)}\left[\pi + \frac{2\pi a^nb^{m}}{(1-a^nb^{m})} \right] \quad \textrm{ from (4)} \\ =& \frac{\pi }{(1-a^2)(1-b^2)}\left[\frac{(1-a^nb^{m})}{(1-a^nb^{m})} + \frac{2 a^nb^{m}}{(1-a^nb^{m})} \right]\\ =& \frac{\pi (1+a^nb^{m})}{(1-a^2)(1-b^2)(1-a^nb^{m})}\\ \end{align*} Hence \[\boxed{ \displaystyle \int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} =\frac{\pi (1+a^nb^{m})}{(1-a^2)(1-b^2)(1-a^nb^{m})} } \]

Appendix

\[ 1+2\sum_{k=1}^{\infty} p^k\cos(kx) = \frac{1-p^2}{1-2p\cos x + p^2} \quad |p|\lt 1 \] Proof \begin{align*} 1+2\sum_{k=1}^{\infty} p^k\cos(kx) =& 1+2\Re\left(\sum_{k=1}^{\infty} p^ke^{ixk}\right)\\ =& 1+2\Re\left(\frac{pe^{ix}}{1-pe^{ix}}\right)\\ =& 1+2\Re\left(\frac{p\cos x +ip\sin x }{1-p\cos x -ip\sin x}\right)\\ =& 1+2\Re\left(\frac{p\cos x +ip\sin x}{1-p\cos x -ip\sin x}\frac{1-p\cos x +ip\sin x}{1-p\cos x+ip\sin x}\right)\\ =& 1+2\Re\left(\frac{p\cos x -p^2 +i(p^2\cos x \sin x +p\sin x (1-p\cos x))}{(1-p\cos)^2+ p^2\sin^2 x}\right)\\ =& 1+\frac{2p\cos x-2p^2}{1-2p\cos x + p^2}\\ =& \frac{1-2p\cos + p^2}{1-2p\cos + p^2}+\frac{2p\cos x-2p^2}{1-2p\cos + p^2}\\ =& \frac{1-p^2}{1-2p\cos + p^2}\\ \end{align*}