Evaluate
∫+∞0logx1+exdx.
I have tried using Feynman's Trick (in several ways, but for example by introducing a variable a such that I(a)=∫+∞0logax1+exdx), but that doesn't seem to work. Also integration by parts and all kinds of substitutions make things worse (I have no idea how to substitute such that log and exp both become simpler.
(Source: Dutch Integration Championship 2013 - Level 5/5)
Answer
By the inverse Laplace transform
∑n≥1(−1)n+1ns=1Γ(s)∫+∞0xs−1ex+1dx
and by differentiating both sides with respect to s
∑n≥1(−1)nlognns=−Γ′(s)Γ(s)2∫+∞0xs−1ex+1dx+1Γ(s)∫+∞0xs−1log(x)ex+1dx
so by evaluating at s=1
∫+∞0logxex+1dx=∑n≥1(−1)nlognn+Γ′(1)⏟−γ∫+∞0dxex+1⏟log2
and it just remains to crack the mysterious series ∑n≥1(−1)nlognn. On the other hand by Frullani's integral, the inverse Laplace transform or Feynman's trick we have log(n)=∫+∞0e−x−e−nxxdx, so
∑n≥1(−1)nlognn=∫+∞0log(1+e−x)−e−xlog2xdx=γlog(2)−12log2(2)
where the last identity follows from the integral representation for the Euler-Mascheroni constant, got by applying the inverse Laplace transform to the series definition γ=∑n≥1[1n−log(1+1n)]. Summarizing, we simply have
∫+∞0log(x)ex+1dx=−12log2(2).
It is possible to prove the equality between the LHS and the RHS of (J) by summation by parts and Euler sums, too.
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