Evaluate ∫01(1ln⁡x+11−x)2dx

David Lewis

David Lewis

Answered

2022-01-06

Evaluate 01(1lnx+11x)2dx

Answer & Explanation

Jeffery Autrey

Jeffery Autrey

Expert

2022-01-07Added 35 answers

Here is an another approach:
Let I denote the integral. By the substitution x=et, we have
I=0[1(1et)22t(1et)+1t2]etdt
=0[et(et1)21t2]dt+0[1+ett22t(et1)]dt
It is easy to observe that the first integral is
0[et(et1)21t2]dt=[1t1et1]0=12
We thus focus on the second integral. Associated to it, we introduce
F(s)=0[1+ett22t(et1)]estdt
By the twice differentiation, we have
F(s)=0[1+et2tet1]estdt
=1s+1s+12n=11(n+s)2
=1s+1s+12ψ1(s+1)
Integrating and using the condition F(+)=0, we have
F(s)=logs+log(s+1)2ψ0(s+1)
Here we used the estimate ψ0(s)logs as s. Integrating again, we have
Suhadolahbb

Suhadolahbb

Expert

2022-01-08Added 32 answers

A bit late to the party, but here is a slightly different approach that doesn't use any series, Stirling's approximations or anything, though does make use of some values of the zeta function.
First substitute x=ey to get
0(11ey)1y)2eydy
The integrand is awkward because when you multiply it out, the pieces separately diverge. Multiplying it by yz regulates the singularity at 0 but also gives standard integrals. So for Re(z)>1 we have (by parts)
01(1ey)2eyyzdy=z0ey1eyyz1dy
=zζ(z)Γ(z)
and
01y(1ey}eyyzdy=ζ(z)Γ(z)
01y2eyyzdy=Γ(z1)
Putting the bits together:
0(11ey1y)2eyyzdy
=Γ(z)(1z1+(z2)ζ(z))
The derivation was valid for Re(z)>1, but both sides of the above equation are analytic for Re(z)1 (RHS has a removable singularity at z=0), so by analytic continuation we may take the limit z0. Near z=0,Γ(z)=1z+O(1), and the final answer is the derivative of 1z1+(z2)ζ(z) at o, i.e.,
1+ζ(0)2ζ(0)=log(2π)32
nick1337

nick1337

Expert

2022-01-11Added 573 answers

OK, I'm going to lay this out up to a sum, which will likely evaluate into whatever answer was provided above. This integral is subject to the same sorts of tricks that I did for another integral involving a factor of 1/logx in the integral. The first piece is to let x=ey; the integral becomes
0dyey((ey(1y))2y2(1ey)2)
Now Taylor expand the factor (1ey)2, and if we can reverse the order of summation and integration, we get:
k=1k0dy(ey(1y))2y2)eky
The integral inside the sum is a bit difficult, although it is convergent. The way I see through it is to replace k with a continuous parameter α and differentiate with respect to α inside the integral twice (to clear the pesky y2 in the denominator) to get a function
I(α)=0dy((ey(1y))2y2)eαyd2Idα2=0dy(ey(1y))2eαy=1α+22α+1+2(α+1)2+1α2α2+2α3
You integrate this twice to recover α; the constants of integration may be shown to vanish by considering the limit as a. The original integral is then
k=1kI(k)
where
I(k)=(k+2)log[k(k+2)(k+1)2]+1k
so the integral takes on the value
k=1[1+[(k+1)21]log(11(k+1)2)]=k=1[1+(k+1)2log(11(k+1)2)]+log2
The sum may be simplified by Taylor expanding the log term; note that the unit value cancels and we get that the integral equals
log2+k=2[1k2m=11m(1k2)m]=log2m=11m+1[ζ(2m)1]
I have not yet evaluated this sum yet, but unless someone else does it before me, I will figure it out and come back.

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?