Why is ∫∞0lnx1+x2dx=0\int_{0}^{\infty} \frac {\ln x}{1+x^2} \mathrm{d}x =0?

We had our final exam yesterday and one of the questions was to find out the value of:
0lnx1+x2dx
Interestingly enough, using the substitution x=1t we get – 10lnx1+x2dx=1lnx1+x2dxand therefore 0lnx1+x2dx=0

I was curious to know about the theory behind this interesting (surprising even!) example.

Thank you.

Answer

When I see an 1+x2 in the denominator it’s tempting to let θ=arctan(x) and dθ=11+x2dx. When you do that here the integral becomes
π20ln(tan(θ))dθ
=π20ln(sin(θ))dθπ20ln(cos(θ))dθ
The two terms cancel because cos(θ)=sin(π2θ).

Also, if you do enough of these, you learn that doing the change of variables from x to 1x converts a dx1+x2 into dx1+x2, so it becomes one of the “tricks of the trade” for integrals with 1+x2 in the denominator. An example: show this trick can be used to show that the following integral is independent of r:
0dx(1+x2)(1+xr)

Attribution
Source : Link , Question Author : Amihai Zivan , Answer Author : Zarrax

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