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This question already has an answer here:

I, I am trying solve the following integral $$\int_{0}^{\infty} \frac{\log^2(x)}{1+x^2}$$

Teachers teached me that I can solve the integral $$\int_{0}^{\infty} \frac{\log^2(x)}{1+x^2}=\frac{d^2}{d\lambda^2} \int_{0}^{\infty} \frac{x^{\lambda}}{1+x^2}$$

I get to the second integral that $$ \int_{0}^{\infty} \frac{x^{\lambda}}{1+x^2}=\frac{2\pi i}{1-e^{2\pi i \lambda}}sin\left(\lambda \frac{\pi}{2}\right)=\frac{-\pi}{e^{\pi i \lambda}sin(\pi\lambda)}\sin\left(\lambda \frac{\pi}{2}\right)$$

so when I try derivate I can get a exist solution because when $\lambda=0$ then $sin(\lambda \pi)=0$

I have to solve the integral by residue methods, with the formula

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marked as duplicate by colormegone, 3SAT, user296602, Ron Gordon complex-analysis Feb 7 '16 at 21:21

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  • $\begingroup$ i love you @RecklessReckoner $\endgroup$ – user311551 Feb 7 '16 at 21:09
  • $\begingroup$ i can not delete this $\endgroup$ – user311551 Feb 7 '16 at 21:10
  • $\begingroup$ @user311551 : work first on $\int_a^b$ and see what happens when $]a;b[ \to ]0;\infty[$ (and you'll probably have to use a theorem for the invertion of two limits because of the derivation) $\endgroup$ – reuns Feb 7 '16 at 21:15