Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
Some users suggested me that I can use Frullani's theorem:
$$\int_0^\infty \frac{f(ax) – f(bx)}{x} = \big[f(0) – f(\infty)\big]\ln \left(\frac ab\right)$$
So I tried to work with that way.
\begin{align} I&=\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}\\ &=\int_0^\infty \frac{\ln \left( x^2+2kx\cos b+k^2\right)-\ln \left( x^2+2kx\cos a+k^2\right)}{x}\mathrm dx\tag{1}\\ &=\int_0^\infty \frac{\ln \left( 1+\dfrac{2k\cos b}{x}+\dfrac{k^2}{x^2}\right)-\ln \left( 1+\dfrac{2k\cos a}{x}+\dfrac{k^2}{x^2}\right)}{x}\mathrm dx\tag{2}\\ \end{align}
The issue arose from $(1)$ because $f(\infty)$ diverges and the same issue arose from $(2)$ because $f(0)$ diverges. I then tried to use D.U.I.S. by differentiating w.r.t. $k$, but it seemed no hope because WolframAlpha gave me this horrible form. Any idea? Any help would be appreciated. Thanks in advance.
Best Answer
First note that by substituting $x\mapsto kx$, we get $$ \int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x} =\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x\cos(b)+1}\right)\frac{\mathrm{d}x}{x} $$ Let $u=\frac{x+\cos(a)}{\sin(a)}$. Then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x} &=-2\int_0^\infty\frac{\sin(a)}{x^2+2x\cos(a)+1}\,\mathrm{d}x\\ &=-2\int_0^\infty\frac{\sin(a)}{(x+\cos(a))^2+\sin^2(a)}\,\mathrm{d}x\\ &=-\frac2{\sin(a)}\int_0^\infty\frac1{\frac{(x+\cos(a))^2}{\sin^2(a)}+1}\,\mathrm{d}x\\ &=-2\int_{\cot(a)}^\infty\frac1{u^2+1}\,\mathrm{d}u\\[9pt] &=-2a \end{align} $$ Integrating in $a$ gives $$ \int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x} =-a^2 $$ Therefore, by subtraction, $$ \int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x} =b^2-a^2 $$