We had our final exam yesterday and one of the questions was to find out the value of:
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.
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
The two terms cancel because cos(θ)=sin(π2−θ).
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: