Sunday, 25 August 2013

definite integrals - Closed form inti0nftyleft(fractanhxx2frac1xe2xright)dx=12logAfrac43log2




Evaluate



0(tanhxx21xe2x)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
12logA43log2



Answer



Let I(z)=za0(12zat+1eat/z1)1eatt2dt
Then Binet's first formula says
I(z)=z0[lnΓ(x)(z12)lnz+zln(2π)2]dx






Letting a=2,z=1/2 gives
4I(12)=0(1214t+1e4t1)1e2tt2dt
and a=4,z=1 gives

4I(1)=0(1214t+1e4t1)1e4tt2dt



Some algebraic manipulation yields
4I(1)4I(12)=0[12t2e2t2t(1214t)(e2te4tt2)]dtJI2



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)dx1/20lnΓ(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(1z)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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