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#1 h Dec 9, 2023, 8:03 AM • 1 Y

Y by Zfn.nom-_nom

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$\int_{0}^{\infty} \log \left(\frac{1+a \sin^{2} bx}{1-a \sin^{2} bx} \right) \frac{1}{x^{2}} \ dx = \frac{\pi b}{2} \left( \sqrt{1+a} - \sqrt{1-a} \right)$

This post has been edited 2 times. Last edited by Martin.s, Dec 9, 2023, 8:03 AM

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#2 h Dec 9, 2023, 11:13 AM • 5 Y

Y by Martin.s, Tip_pay, Talker, Zfn.nom-_nom, MS_asdfgzxcvb

$I(a,b)=\int_{0}^{\infty} \log \left(\frac{1+a \sin^{2} bx}{1-a \sin^{2} bx} \right) \frac{1}{x^{2}} \ dx =b\int_{0}^{\infty} \log \left(\frac{1+a \sin^{2} x}{1-a \sin^{2} x} \right) \frac{1}{x^{2}} \ dx$ As $\sin^2x$ is periodic with the period $\pi$ , we can apply Lobachevsky integral formula
$I(a,b)=b\int_0^{\pi/2} \log \left(\frac{1+a \sin^{2} x}{1-a \sin^{2} x} \right)\frac{dx}{\sin^2x}=b\int_0^{\pi/2} \log \left(\frac{1+a \cos^{2} x}{1-a \cos^{2} x} \right)\frac{dx}{\cos^2x}\overset{t=\tan x}{=}b\int_0^\infty\ln\left(\frac{1+t^2+a}{1+t^2-a}\right)dt$ $\overset{IBP}{=}2b\int_0^\infty\left(\frac1{1+t^2+a}-\frac1{1+t^2-a}\right)t^2dt\overset{t^2=x}{=}2ab\int_0^\infty\frac{\sqrt x}{(1+x)^2-a^2}dx=2ab\int_0^\infty\frac{\sqrt x}{(x+1+a)(x+1-a)}dx$ Using the keyhole contour in the complex plane and evaluating residues (counter-clockwise, at $z=e^{\pi i}(1+a);\,z=e^{\pi i}(1-a)\,$ )
$2I=2\pi i\cdot2ab\Big(\frac1{-2a}\sqrt{1+a}e^{\frac{\pi i}2}+\frac1{2a}\sqrt{1-a}e^{\frac{\pi i}2}\Big)\,\,\Rightarrow\,\,\boxed{\,\,I(a,b)=\pi b\Big(\sqrt{1+a}\,-\,\sqrt{1-a}\Big)\,\,}$

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