Evaluate \int_0^1(\frac{1}{\ln x}+\frac{1}{1-x})^2dx

David Lewis

David Lewis

Answered question

2022-01-06

Evaluate 01(1lnx+11x)2dx

Answer & Explanation

Jeffery Autrey

Jeffery Autrey

Beginner2022-01-07Added 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

Beginner2022-01-08Added 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

Expert2022-01-11Added 777 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.

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