Some time ago I came across to the following integral: I=∫011−x1+xdxln⁡x What are the hints...

David Young

David Young

Answered

2022-01-04

Some time ago I came across to the following integral:
I=011x1+xdxlnx
What are the hints on how to compute this integral?

Answer & Explanation

esfloravaou

esfloravaou

Expert

2022-01-05Added 43 answers

Make the substitution x=ey and do a little algebra to get the value of the integral to be
0dyyeye2y1+ey
Now Taylor expand the denominator and get
0dyy(eye2y)k=0(1)keky
If we can reverse the order of sum and integral, we get
k=0(1)k0dyy(e(k+1)ye(k+2)y)
The integral inside may be evaluated exactly, and the result is the sum
k=0(1)klogk+1k+2
=limnlog12342n12n23562n+12n+2
=log(2π)
jgardner33v4

jgardner33v4

Expert

2022-01-06Added 35 answers

I tried with Differentiation under integration sign:
J(α)=011xα1+xdxlnx α>0
Then as usual,
dJdα=01xα1+xdx=f(α)
Then, integrating
J(α)=f(α)dα+C
I used Mathematica to evaluate f(α) (it gives difference of two Harmonic numbers) and its integral. The integral is just lnΓ(2m+12)Γ(m+12). Note that J(0)=0 and putting α=1, the result follows.
nick1337

nick1337

Expert

2022-01-11Added 573 answers

Consider
I(α)=011xα(1+x)lnxdx (1)
Differentiating (1) with respect α yields
dIdα=01ddα[1xα(1+x)lnx]dx=01xα1+xdx=01k=0(1)kxα+kdx=k=0(1)k01xα+kdx=k=0(1)kα+k+1 (2)
Now consider polygamma function
ψn(z)=dn+1dzn+1lnΓ(z) (3)
and
ψn(z)=(1)n+1n!k=01(z+k)n+1 (4)
Hence by using (4) we obtain
k=0(1)k(z+k)n+1=1(2)n+1n![ψn(z2)ψn(z+12)] (5)
Using (3) and (5) then (2) becomes
dIdα=12[ψ0(α+12)ψ0(α+22)]I(α)=12[ψ0(α+12)ψ0(α+22)]dα=lnΓ(α+12)lnΓ(α+22)+C=lnΓ(α+12)lnΓ(α+22)+12lnπwhere I(0)=0 and C=lnΓ(12)Thus011x(1+x)lnxdx=I(1)=lnΓ(1)lnΓ(32)12lnπ=ln(2π)

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