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Resolve $$ \int_{0}^{\infty} \dfrac{(\ln(x))^2}{1+x^2} \, dx $$ For this, we have to consider $ log_{-\pi/2}(z) $ and the curve of integration $ \gamma_{\varepsilon,R} $ formed by the next curves: a) $ \gamma_{1}(t) = t $ with $ t \in [\varepsilon,R] $, b) $ \gamma_{2}(t) = R e^{it} $ with $ t \in [0,\pi] $, c) $ \gamma_{3}(t) = t $ with $ t \in [-R,-\varepsilon] $, d) $ \gamma_{4}(t) = \varepsilon e^{it} $ with $ t \in [0,\pi] $ with the other orientation. Here, $ 0 < \varepsilon < R < 1 $.

Тyma Gaidash
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José Carlos Pérez Garrido
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  • You can consider instead the equivalent integral: $$\int_{0}^{\infty}\frac{x^2}{\cosh x},\mathrm{d}x$$Which is probably easier. Use $x\mapsto e^x$ and the evenness of cosh. – FShrike Dec 26 '21 at 17:23
  • Please show what happened when you tried those curves. – J.G. Dec 26 '21 at 17:24
  • You must use this curves to resolve this integral, you can not do it with other method. – José Carlos Pérez Garrido Dec 26 '21 at 17:31
  • I think you meant $0<\varepsilon<1<R$. – J.G. Dec 26 '21 at 17:36
  • I guess the notation $\log_{-\pi/2}$ denotes a branch of the natural logarithm with cut along the negative imaginary axis. I agree with @J.G. ... you should show us more of your attempt. Maybe begin by explaining the outline of what you should do (before you begin doing any computations). – GEdgar Dec 26 '21 at 17:40
  • For reference, using the geometric series suggested by my form of the integral and the Gamma function, one finds the answer to be: $$4\sum_{n\ge1}\frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{8}$$Reference – FShrike Dec 26 '21 at 17:41
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    Integrate $\frac{\ln^2z}{1+z^2}$ around $\gamma_{\varepsilon,,R}$. Explain why b) $\to0$ as $R\to\infty$ and d) $\to0$ as $\varepsilon\to0^+$, so$$\begin{align}2\pi i\lim_{z\to i}\frac{\ln^2z}{z+i}&=\int_0^\infty\frac{\ln^2xdx}{1+x^2}+\int_{-\infty}^0\frac{\ln^2xdx}{1+x^2}\&=\int_0^\infty\frac{(\ln^2x+(\ln x+i\pi)^2)dx}{1+x^2}\&=2\int_0^\infty\frac{\ln^2xdx}{1+x^2}-\frac12\pi^3+2i\pi\int_0^\infty\frac{\ln xdx}{1+x^2}.\end{align}$$Prove $\int_0^\infty\frac{\ln xdx}{1+x^2}=0$ and finish the arithmetic. – J.G. Dec 26 '21 at 18:09

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