Let γ be the Euler-Mascheroni constant.
I'm trying to prove that
2∫∞0ex−x−1x(e2x−1)dx=ln(π)−γ
I tried introducing a parameter to the exponent in the numerator and then differentiating under the integral sign. But doing so seems to result in an integral that doesn't converge.
Answer
Hint. One may set
$$
f(s):=2\int_0^\infty \frac{e^{sx}-sx-1}{x(e^{2x}-1)}dx, \quad 0Inordertogetridofthefactor$x$inthedenominator,wemaydifferentiateundertheintegralsigngetting
f'(s)=2\int_0^\infty \frac{e^{sx}-1}{e^{2x}-1}dx, \quad 0Thenexpandingthelatterintegrandin$e−kx$termsandintegratingtermwiseweget
f'(s)=-\gamma-\psi\left(1-\frac{s}2\right) \tag3
$$ where ψ:=Γ′/Γ and where γ is the Euler-Mascheroni constant.
Integrating (3), with the fact that, as s→0, f(s)→0, we get
$$
2\int_0^\infty \frac{e^{sx}-sx-1}{x(e^{2x}-1)}dx=-\gamma s+2 \log \Gamma\left(1-\frac{s}2\right), \quad 0$$
from which you deduce the value of your initial integral by putting s:=1.
Identity (4) is much more than what was asked.
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