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

Zerrilloh6

Zerrilloh6

Answered question

2022-01-03

How to solve 0log(x+1x)1+x2dx ?

Answer & Explanation

rodclassique4r

rodclassique4r

Beginner2022-01-04Added 37 answers

You may write
0log(x+1x)1+x2dx=0log(1+x2)logx1+x2dx
=0log(1+x2)1+x2dx0logx1+x2dx
Clearly, by the change of variable 1x, we get
=0logx1+x2dx=0logx1+x2dx
On the other hand, by the change of variable
x=tanθ,dx=(1tan2θ)dθ,1+tan2θ=1cos2θ
we obtain the classic evaluation:
=0log(1+x2)1+x2dx=20π2log(cosθ)dθ=πlog2

Hector Roberts

Hector Roberts

Beginner2022-01-05Added 31 answers

Instead of using the trigonometric subsitution you may use the identity
log(1+x2)=01dax21+ax2
Calling your integral of interest I It follows that
I=01da0dxx2(x2+1)(1+ax2)use Residue Theorem to calculate this integral
=01daπ2(a+a)use  a=b2  to turn this into a standard integra=πlog(2)
In agreement with Olivier’s answer
Vasquez

Vasquez

Expert2022-01-11Added 669 answers

xtanxI=0π2log(sinxcosx)dx=0pi2log(sinx)dx+0π2log(cosx)dx=20π2log(cosx)dx=πlog2

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