This integral popped up recently
∫∞−∞e2x−exx(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=2∫∞0e2x−exxe3x(1+e−2x)(1+e−x)dx=2∫∞0e−2x−e−xx(∞∑n=0(−1)ne−nx)(∞∑m=0(−1)me−2mx)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
e2x−ex(1+e2x)(1+ex)=11+ex−11+e2x=−∫21ddu(11+eux)⋅du,
that is,
e2x−exx(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
∫∞−∞e2x−exx(1+e2x)(1+ex)dx=∫21duu=log2.
More generally, for every positive a and b,
∫∞−∞eax−ebxx(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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