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How would one show that $$\int_0^{\infty}\frac{\ln^2 x}{(1+x)(1+xe^{-2\pi})(1+xe^{-4\pi})\cdots}\, dx=\pi^3\left(1+\frac{1}{\pi}+8\left(\frac{1}{2}-\frac{1}{e^{2\pi}-1}-\frac{1}{e^{4\pi}-1}-\frac{1}{e^{6\pi}-1}-\cdots\right)^2\right)$$ ?

It is shown in this instagram post but the author did not show any work. I checked in Wolfram alpha and the numerical values match, but I couldn't prove it. It looks like that one integral from Ramanujan but with a $\ln^2(x)$ on the integrand, so I tried differentiating $x^{s-1}$. But the integral soon get messy and I couldn't proceed. I am now thinking an alternative way to prove this identity.

Any help is appreciated.

Edit: Using the technique in this answer probably works

Brightsun
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1 Answers1

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Too long for a comment.

It is not so hard if you perform partial fraction decomposition $$I_n=\int_0^t \frac{\log ^2(x)}{\left(-x;e^{-2 \pi }\right){}_{n+1}}\,dx=\int_0^t \sum_{k=0}^\infty A_k\frac{\log ^2(x)}{1+x\,e^{-2 \pi k} }\,dx$$

$$J_k=\int_0^t \frac{\log ^2(x)}{1+x\,e^{-2 \pi k} }\,dx$$ $$J_k=e^{2 \pi k} \left(-2 \text{Li}_3\left(-e^{-2 k \pi } t\right)+2 \log (t) \text{Li}_2\left(-e^{-2 k \pi } t\right)+\log ^2(t) \log \left(e^{-2 \pi k} t+1\right)\right)$$

Recombine everything and use the limit for $t\to \infty$.

$$I_1=\frac{5}{3} \pi ^3 (1+\coth (\pi ))$$ $$I_2=\frac{1}{6} e^{\pi } \left(5 e^{2 \pi }-29\right) \pi ^3 \text{csch}(\pi ) \text{csch}(2\pi )$$ $$I_3=\frac{e^{3 \pi } \pi ^3 \text{csch}^3(\pi ) (-29+41 \cosh (3 \pi ) \text{sech}(\pi )-36 \sinh (3 \pi ) \text{sech}(\pi ))}{12+24 \cosh (2 \pi )}$$

The problem I face is to switch to the kind of expression you give in the post.

Edit

If $n\to \infty$, for the rhs of the given expression, notice that $$\sum_{1}^\infty \frac{1}{e^{2 \pi k}-1}=-\frac{\psi _{e^{-2 \pi }}^{(0)}(1)+\log \left(1-e^{-2 \pi }\right)}{2 \pi }=1.87443\times 10^{-3}$$

Claude Leibovici
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