Evaluate
∫∞0(tanhxx2−1xe2x)dx
I haven't been able to find references to the indefinite integral of the tanh term except for some similar forms that had solutions. see here and here.
Edit:
Following Random Variable's result we have the form
12logA−43log2
Answer
Let I(z)=za∫∞0(12−zat+1eat/z−1)1−e−att2dt
Then Binet's first formula says
I(z)=∫z0[lnΓ(x)−(z−12)lnz+z−ln(2π)2]dx
Letting a=2,z=1/2 gives
4I(12)=∫∞0(12−14t+1e4t−1)1−e−2tt2dt
and a=4,z=1 gives
4I(1)=∫∞0(12−14t+1e4t−1)1−e−4tt2dt
Some algebraic manipulation yields
4I(1)−4I(12)=∫∞0[12t2−e−2t2t−(12−14t)(e−2t−e−4tt2)]dt⏟J−I2
With I your desired integral. Surprisingly, J has elementary primitive (even it were not, we still have systematic way to crash it), with value 5/8.
Hence it remains to evalute ∫10lnΓ(x)dx∫1/20lnΓ(x)dx
The former is just ln(2π)/2, for the latter, we can use the integral representation of Barnes G function:
∫z0lnΓ(x)dx=z(1−z)2+z2ln(2π)+zlnΓ(z)−logG(1+z)
and the special value lnG(32)=−32lnA+lnπ4+18+ln224
with A being the Glaisher-Kinkelin constant.
Alternatively, use Fourier expansion of lnΓ(x), integrate termwise, and remember the relation between A and ζ′(2) also gives the value of the integral ∫1/20lnΓ(x)dx=32lnA+524ln2+lnπ4
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