Sunday, 18 December 2016

integration - Prove that intinftyinftyfrace2xexx(1+e2x)(1+ex)mathrmdx=log2



This integral popped up recently



e2xexx(1+e2x)(1+ex)dx=log2



A solution using both real and complex analysis is welcome. I tried rewriting it using symmetry, and then the series expansion of 1/(1+enx), however this did not quite make it.



I=20e2xexxe3x(1+e2x)(1+ex)dx=20e2xexx(n=0(1)nenx)(m=0(1)me2mx)dx


The first part reminds me of a Frullani integral (it evaluates to log2). However I am unsure if this is the correct path, any help would be appreciated. =)


Answer



One can start from the identity
e2xex(1+e2x)(1+ex)=11+ex11+e2x=21ddu(11+eux)du,


that is,

e2xexx(1+e2x)(1+ex)=21eux(1+eux)2du.

Note that, for every nonzero u,
eux(1+eux)2dx=1u(1+eux)|=1|u|,

hence Fubini theorem shows that the integral to be computed is
e2xexx(1+e2x)(1+ex)dx=21duu=log2.

More generally, for every positive a and b,
eaxebxx(1+eax)(1+ebx)dx=log(ab).

This is rediscovering the fact that, for every monotonous function φ with limits at ±,
Rφ(ax)φ(bx)xdx=(φ(+)φ())log(ba).


in the present case for the function
φ(x)=11+ex.


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