What is the value of $$\frac{1}{\pi^{2}} \int_{0}^{\infty} \frac{(\ln x)^{2}}{\sqrt{x}(1-x)^{2}} d x$$
I tried with by parts.But not able to Proceed further. Tried with put $x=\sin^2\theta$
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FDP
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Math_centric
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I tried with your trigo part, but nothing is happening – maveric Mar 28 '20 at 17:36
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I'm pretty sure that diverges at $0$. – B. Goddard Mar 28 '20 at 17:40
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Can we integrate it? – maveric Mar 28 '20 at 17:41
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$x=\sin^2\theta$ is not a valid substitution because $x\in(0,\infty)$ but $\sin^2\theta$ can never go that far. – Ninad Munshi Mar 28 '20 at 17:53
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1@B.Goddard That is incorrect. The singularity at $0$ is an integrable one, so it converges. – Ninad Munshi Mar 28 '20 at 17:55
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$\displaystyle \int_0^\infty (...)=\int_0^1 (...)+\int_1^\infty (...)$ then perform the change of variable $y=1/x$ in the second integral. – FDP Mar 28 '20 at 17:56
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Wolfram Alpha gives a result of 2. – sammy gerbil Mar 28 '20 at 17:57
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@FDP can u give soln – maveric Mar 28 '20 at 18:08
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You just need to let $x\to t^2$ to arrive at the duplicated integral. – Zacky Mar 28 '20 at 18:16