A math contest problem ∫10ln(1+ln2x4π2)ln(1−x)x dx\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx

A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.

Prove:
10ln(1+ln2x4π2)ln(1x)xdx=π2(4ζ(1)+23).

Answer

Here is a solution: Let I denote the integral. Then

I=10log(1+log2x4π2)Li2(x)dx=[log(1+log2x4π2)Li2(x)]10+210logx4π2+log2xLi2(x)xdx=20t4π2+t2Li2(et)dt(x=et)=2n=11n20t4π2+t2entdt=2n=11n20(0cos(2πu)etudu)entdt=2n=11n20cos(2πu)u+ndu=2n=11n20cos(2πnu)u+1du(unu)=2n=1k=11n210cos(2πnu)u+kdu=2k=1101u+k(n=1cos(2πnu)n2)du.

Now we invoke the Fourier series of the Bernoulli polynomial B2(x):

n=1cos2πnxn2=π2B2(x)=π2(x2x+16),0x1.

Then it follows that

I=2π2k=110u2u+16u+kdu=π2k=1{2k+1+2(k2+k+16)log(kk+1)}.

Now we consider the exponential of the partial sum:

exp[Nk=1{2k+1+2(k2+k+16)log(kk+1)}]=eN2+2NNk=1(kk+1)2k2+2k+13=eN2+2N(N+1)2N2+2N+13Nk=1k4k=e2N(1+1N)2N2+2N+13{eN2/4NN2/2+N/2+1/12Nk=1kk}4.

In view of the definition of Glaisher-Kinkelin constant A, we have

\lim_{N\to\infty} \exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] = \frac{A^{4}}{e}.

This, together with the identity \log A = \frac{1}{12} – \zeta'(-1), yields

I = \pi^{2} ( 4 \log A – 1 ) = -\pi^{2} \left(4\zeta'(-1) + \frac{2}{3} \right)

as desired.

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Bennett Gardiner

Leave a Comment