Integral of the day II

Integral related to the Sinc function

We show the proof of this mysterious integral:
\[\int_{0}^{\infty}\frac{\sin x}{x} \ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) = \pi \ln\left( \frac{\sqrt{a^2+1}+a}{\sqrt{b^2+1}+b}\right) \quad |a|,|b|> 1\]
The integral resisted various techniques (residues, laplace transform) but following a hint suggested by @mvs_rpi we managed to obtain the proof: It turned out that this integral is a Lobachevsky integral.

The Lobachevsky integral formula states that:

If $f(x)$ is a continous function satisfying the $\pi$-periodic assumption $f(x+\pi) = f(x)$, and $f(\pi-x)=f(x)$, for $0\leq x <\infty$

then
\[\int_{0}^{\infty} \frac{\sin^2 x}{x^2}f(x) dx = \int_{0}^{\infty} \frac{\sin x}{x} f(x)dx = \int_{0}^{\frac{\pi}{2}} f(x) dx \]
Clearly
\[f(x) = \ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right)\]
is a $\pi-$ periodic function.

Therefore we just have to calculate
\[ \int_{0}^{\frac{\pi}{2}}\ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) dx \]
If we split the integral in two:
\[ \int_{0}^{\frac{\pi}{2}}\ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) dx = \underbrace{\int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx}_{I_{1}} - \underbrace{\int_{0}^{\frac{\pi}{2}} \ln(b^2+\cos^2 x)dx}_{I_{2}} \]
if we find $I_{1}$ automatically we also have $I_{2}$


Proof:

First, we will show that \[ \int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx = \pi \ln(\sqrt{a^2+1} +a)-\pi\ln(2) \quad |a|>1 \] \begin{align*} I_{1} = \int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx =& \int_{0}^{\frac{\pi}{2}} \ln\left(a^2\left(1+\frac{1}{a^2}\cos^2 x\right)\right)dx\\ =& \int_{0}^{\frac{\pi}{2}}\ln(a^2)dx + \int_{0}^{\frac{\pi}{2}}\ln\left(1+\frac{1}{a^2}\cos^2 x\right)dx\\ =& \pi\ln(a) + \int_{0}^{\frac{\pi}{2}}\sum_{k=1}^{\infty} \left[\frac{(-1)^{k+1}}{k} \frac{\cos^{2k}x}{a^{2k}} \right] dx \quad (|a|>1)\\ =& \pi\ln(a) + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{ka^{2k}} \int_{0}^{\frac{\pi}{2}}\cos^{2k} x dx\\ =& \pi\ln(a) - \sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \int_{0}^{\frac{\pi}{2}} w^{2k}(1-w^2)^{-\frac{1}{2}} dw \quad (w \mapsto \cos x)\\ =& \pi\ln(a) - \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \int_{0}^{\frac{\pi}{2}} s^{k+\frac{1}{2}-1}(1-s)^{\frac{1}{2}-1} ds \quad (s \mapsto w^2)\\ =& \pi\ln(a) - \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k B\left(k+\frac{1}{2}, \frac{1}{2}\right)\\ =& \pi\ln(a) - \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(k+1)}\\ \end{align*} Recall that \[ \Gamma\left(k+\frac{1}{2}\right) = \frac{(2k-1)!!\sqrt{\pi}}{2^k}\] \[ \Gamma(k+1)= \frac{(2k)!!}{2^k} \] Hence \[ \frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma(k+1)} = \frac{(2k-1)!!\sqrt{\pi}}{(2k)!!} \] Therefore \[ I_{1} = \pi\ln(a) - \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \frac{(2k-1)!!}{(2k)!!} \] In another post we proved that \[ \frac{1}{\sqrt{1-x}} = \sum_{k=0}^{\infty} \frac{(2k-1)!!}{(2k)!!} x^k\] Then, substracting $-1$ and dividing by $\displaystyle \frac{1}{x}$ afterwards (the same trick we used in this post): \[\frac{1}{\sqrt{1-x}}-1 = \sum_{k=1}^{\infty} \frac{(2k-1)!!}{(2k)!!} x^k\] \[\Longrightarrow \frac{1}{x\sqrt{1-x}}-\frac{1}{x} = \sum_{k=1}^{\infty} \frac{(2k-1)!!}{(2k)!!} x^{k-1}\] Integrating: \[\int_{0}^{x}\frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = \sum_{k=1}^{\infty} \frac{(2k-1)!!}{(2k)!!} \int_{0}^{x}x^{k-1}dx = \sum_{k=1}^{\infty}\frac{1}{k} \frac{(2k-1)!!}{(2k)!!} x^k\] But note that \[ \int\frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = -2\ln\left(\sqrt{1-x}+1\right)+C\] Therefore \[\sum_{k=1}^{\infty}\frac{1}{k} \frac{(2k-1)!!}{(2k)!!} x^k = -2 \ln\left(\sqrt{1-x}+1\right)+2\ln(2) \] If we put $x = -\frac{1}{a^2}$ \[\sum_{k=1}^{\infty}\frac{1}{k} \frac{(2k-1)!!}{(2k)!!} \left(-\frac{1}{a^2}\right)^k = -2 \ln\left(\sqrt{1+\frac{1}{a^2}}+1\right)+2\ln(2) \] Therefore \begin{align*} I_{1} =& \pi\ln(a) - \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \frac{(2k-1)!!}{(2k)!!} \\ =& \pi\ln(a)+ \pi \ln\left(\sqrt{1+\frac{1}{a^2}}+1\right)-\pi \ln(2)\\ =& \pi\ln(a)+ \pi \ln\left(\sqrt{\frac{a^2+1}{a^2}}+\frac{a}{a}\right)-\pi \ln(2)\\ =& \pi\ln(a)+ \pi \ln\left(\frac{1}{a} (\sqrt{a^2+1}+a)\right)-\pi \ln(2)\\ =& \pi\ln(a) -\pi \ln(a) + \pi \ln\left( (\sqrt{a^2+1}+a)\right)-\pi \ln(2)\\ =& \pi \ln\left( \sqrt{a^2+1}+a\right)-\pi \ln(2) \end{align*} Therefore \[I_{1}= \int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx = \pi \ln\left( \sqrt{a^2+1}+a\right)-\pi \ln(2)\] \[I_{2}= \int_{0}^{\frac{\pi}{2}} \ln(b^2+\cos^2 x)dx = \pi \ln\left( \sqrt{b^2+1}+b\right)-\pi \ln(2)\] Hence, we have the desired result \[ \int_{0}^{\frac{\pi}{2}}\ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) dx = \pi \ln\left( \frac{\sqrt{a^2+1}+a}{\sqrt{b^2+1}+b}\right)\] By the Lobachevsky integral formula: \[\boxed{\int_{0}^{\infty}\frac{\sin x}{x} \ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) = \pi \ln\left( \frac{\sqrt{a^2+1}+a}{\sqrt{b^2+1}+b}\right) \quad |a|,|b|>1} \]