Evaluating the definite integral $\int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1} dx$

Asked Feb 16 '15 at 06:56

Active Feb 16 '15 at 07:07

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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 $$\frac{\pi^3}{8}.$$

edited Jun 12 '20 at 10:38

Community

asked Feb 16 '15 at 06:56

Akkf

@dustin it there any possible solution without using complex analysis? I haven't learnt it yet... – Akkf Feb 16 '15 at 07:11
Yes look at this answer on the same post. – dustin Feb 16 '15 at 07:12
Here's another duplicate – mrf Feb 16 '15 at 08:14

Evaluating the definite integral $\int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1} dx$

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