What are the usual ways to follow in order to solve the integrals given below?...

Salvatore Boone

Salvatore Boone

Answered

2022-01-03

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

Answer & Explanation

poleglit3

poleglit3

Expert

2022-01-04Added 32 answers

As an addendum of sorts to the previous answers, there is the identity
logG(z+1)=z2log(2π)z(z+1)2+zlogΓ(z+1)z(logz1)0zlogΓ(t)dt
where logG(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 as
G(n)=k=1n2k!
Thus, to evaluate 01logΓ(t)dt, we have
logG(2)=12log(2π)1+logΓ(2)(log11)01logΓ(t)dt
0=12log(2π)01logΓ(t)dt
and you obtain the same solution as Andrew.
For the integral 01tlogΓ(t)dt, integration by parts and taking appropriate limits yields the identity
01tlogΓ(t)dt=1201t2ψ(t)dt
Now, Victor Adamchik, in a paper on negative-order polygamma functions (the same sort of functions that appear in Argon's answer), gives the identity
0zxnψ(x)dx=
=(1)n(Bn+1Hnn+1ζ(n))+k=0n(1)k(nk)znk(ζ(k,z)Bk+1(z)Hkk+1
where Bn and
lovagwb

lovagwb

Expert

2022-01-05Added 50 answers

As for the first integral, one can use the Euler's reflection formula Γ(1z)Γ(z)=πsinπz
I=1201(log(Γ(x)+logΓ(1x))dx=1201logπsinπxdx=
=1201(logπlogsinπx)dx=12logπ12π0πlogsinxdx
=12logπ12π(πlog2)=12log2π
The last integral is well known Gauss integral.
karton

karton

Expert

2022-01-09Added 439 answers

As for J, another way is to try to use the Fourier series for lnΓ(x) discovered by E.E. Kummer in 1847:
lnΓ(x)=ln2π2+n=1cos2πnx2n+n=1(γ+ln2πn)sin2πnxnπ(0<x<1)
where γ=0.577... is Euler's constant
Let's multiply this equality by x and integrate from 0 to 1.
Integrals on the right side:
01xdx=12
01xcos2πnxdx=0
01xsin2πnxdx=12πn
Thus,
01xlnΓ(x)=ln2π4γ2π2n=11n212π2n=1ln2πnn2
=ln2π4γ1212π2n=1ln2πnn2
if I am not mistaken. I don't know can this be simplified further.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?