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This question already has an answer here:

I have two integral questions listed below:

$$\int_0^{\infty} \frac{\ln x}{x^2 + 1} dx \qquad (1)$$

$$\int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1} dx \qquad (2)$$

The first one, I've solved it, by separating the integral into integrals over $0$ to $1$ and over $1$ to $\infty$; with substitution, it becomes zero.

But the second one wasn't that easy. Can someone give me help?

Wolfram alpha says the answer is $$\frac{\pi^3}{8}.$$

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marked as duplicate by dustin, Claude Leibovici, TZakrevskiy, mrf, Hans Lundmark Feb 16 '15 at 8:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @dustin it there any possible solution without using complex analysis? I haven't learnt it yet... $\endgroup$ – Akkf Feb 16 '15 at 7:11
  • $\begingroup$ Yes look at this answer on the same post. $\endgroup$ – dustin Feb 16 '15 at 7:12
  • $\begingroup$ Here's another duplicate $\endgroup$ – mrf Feb 16 '15 at 8:14