What are the usual ways to follow in order to solve the integrals given below? I=∫01ln⁡Γ(x)dx J=∫01xΓ(x)dx

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What are the usual ways to follow in order to solve the integrals given below?  I=∫01ln⁡Γ(x)dx  J=∫01xΓ(x)dx

poleglit3 · Accepted Answer

As an addendum of sorts to the previous answers, there is the identitylog⁡G(z+1)=z2log⁡(2π)−z(z+1)2+zlog⁡Γ(z+1)−z(log⁡z−1)−∫0zlog⁡Γ(t)dtwhere log⁡G(z) is the logarithm of the Barnes function (double gamma function) G(z), the function that satisfies the functional equation G(z+1)=Γ(z)G(z). (Barnes proved this identity in his paper, where he introduced the function now named after him.) For n an integer, G(n) can be expressed asG(n)=∏k=1n−2k!Thus, to evaluate ∫01log⁡Γ(t)dt, we havelog⁡G(2)=12log⁡(2π)−1+log⁡Γ(2)−(log⁡1−1)−∫01log⁡Γ(t)dt0=12log⁡(2π)−∫01log⁡Γ(t)dtand you obtain the same solution as Andrew.For the integral ∫01tlog⁡Γ(t)dt, integration by parts and taking appropriate limits yields the identity∫01tlog⁡Γ(t)dt=−12∫01t2ψ(t)dtNow, Victor Adamchik, in a paper on negative-order polygamma functions (the same sort of functions that appear in Argon&#039;s answer), gives the identity∫0zxnψ(x)dx==(−1)n(Bn+1Hnn+1−ζ′(−n))+∑k=0n(−1)k(nk)zn−k(ζ(−k,z)−Bk+1(z)Hkk+1where Bn and

lovagwb · Accepted Answer

As for the first integral, one can use the Euler&#039;s reflection formula Γ(1−z)Γ(z)=πsin⁡πzI=12∫01(log⁡(Γ(x)+log⁡Γ(1−x))dx=12∫01log⁡πsin⁡πxdx==12∫01(log⁡π−log⁡sin⁡πx)dx=12log⁡π−12π∫0πlog⁡sin⁡xdx=12log⁡π−12π(−πlog⁡2)=12log⁡2πThe last integral is well known Gauss integral.

karton · Accepted Answer

As for J, another way is to try to use the Fourier series for ln⁡Γ(x) discovered by E.E. Kummer in 1847: ln⁡Γ(x)=ln⁡2π2+∑n=1∞cos⁡2πnx2n+∑n=1∞(γ+ln⁡2πn)sin⁡2πnxnπ(0&amp;lt;x&amp;lt;1) where γ=0.577... is Euler&#039;s constant Let&#039;s multiply this equality by x and integrate from 0 to 1. Integrals on the right side: ∫01xdx=12 ∫01xcos⁡2πnxdx=0 ∫01xsin⁡2πnxdx=−12πn Thus, ∫01xln⁡Γ(x)=ln⁡2π4−γ2π2∑n=1∞1n2−12π2∑n=1∞ln⁡2πnn2 =ln⁡2π4−γ12−12π2∑n=1∞ln⁡2πnn2 if I am not mistaken. I don&#039;t know can this be simplified further.

What are the usual ways to follow in order to

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