How to solve ∫0∞log⁡(x+1x)1+x2dx ?

Question

rodclassique4r · Accepted Answer

You may write ∫0∞log⁡(x+1x)1+x2dx=∫0∞log⁡(1+x2)−log⁡x1+x2dx =∫0∞log⁡(1+x2)1+x2dx−∫0∞log⁡x1+x2dx Clearly, by the change of variable →1x, we get =∫0∞log⁡x1+x2dx=−∫0∞log⁡x1+x2dx On the other hand, by the change of variable x=tan⁡θ,dx=(1−tan2θ)dθ,1+tan2θ=1cos2θ we obtain the classic evaluation: =∫0∞log⁡(1+x2)1+x2dx=−2∫0π2log⁡(cos⁡θ)dθ=πlog⁡2

Hector Roberts · Accepted Answer

Instead of using the trigonometric subsitution you may use the identity

\log (1 + x^{2}) = \int_{0}^{1} d a \frac{x^{2}}{1 + a x^{2}}

Calling your integral of interest

I

It follows that

I = \int_{0}^{1} d a \underset{use Residue Theorem to calculate this integral}{\underset{⏟}{\int_{0}^{\infty} d x \frac{x^{2}}{(x^{2} + 1) (1 + a x^{2})}}}

= \underset{use a = b^{2} to turn this into a standard integra}{\underset{⏟}{\int_{0}^{1} d a \frac{π}{2 (a + \sqrt{a})}}} = π \log (2)

In agreement with Olivier’s answer

Vasquez · Accepted Answer

x→tan⁡xI=−∫0π2log⁡(sin⁡xcos⁡x)dx=−∫0pi2log⁡(sin⁡x)dx+∫0π2log⁡(cos⁡x)dx=−2∫0π2log⁡(cos⁡x)dx=πlog⁡2

How to solve \int^\infty_0\frac{\log(x+\frac{1}{x})}{1+x^2}dx ?

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