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How to evaluate this? $$\int_0^{\infty}\frac{\ln\left(1+x\right)\ln\left(1+\frac{1}{x^2}\right)}{x}\,dx$$

I was given this integral saying that it can be solved, but when I couldn’t, I gave up and tried to give this input to integral calculator and as it turns out, even it wasn’t able to give an output in terms of any operator. Please help me.

Asv
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  • Maybe the real question is to show that the integral converges. That is an interesting exercise. – GEdgar Jun 11 '20 at 12:44
  • @GEdgar both showing that it converges and evaluating the integral would be very interesting – Asv Jun 11 '20 at 12:50
  • You have a logaritm in the form $log(1+ \alpha)$ so the most obvious thing is to find the McLaurin polynomials of the integrand function for $x\longrightarrow +\infty$ and $x \longrightarrow 0$. – Vajra Jun 11 '20 at 16:21
  • The convolution theorem for the Mellin transform gives $$\int_{\mathbb R^+} x^{-1}\ln(1 + x) \ln(1 + x^{-2}) , dx = \mathcal M^{-1}F(p) F(p/2)/2, \ F(p) = \mathcal Mx \mapsto \ln(1 + x) = \pi p^{-1} \csc \pi p, \quad -1 < \operatorname {Re} p < 0.$$ The residues at $p = -2 k$ and $p = -2 k + 1$ give sums of $(-1)^k/(2 k^3)$ and $(-1)^{k - 1} \pi /(2 k - 1)^2$, which are not elementary. – Maxim Jun 11 '20 at 17:04
  • possible duplicate here https://math.stackexchange.com/questions/730732/integral-i-int-0-infty-frac-ln1x-ln1x-2x-dx/3447541#3447541 – Ali Shadhar Jun 12 '20 at 20:08

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$$I=\int_0^{1}\frac{\ln\left(1+x\right)\ln\left(1+\frac{1}{x^2}\right)}{x}\,dx+\int_1^{\infty}\frac{\ln\left(1+x\right)\ln\left(1+\frac{1}{x^2}\right)}{x}\,dx$$ $$\int_1^{\infty}\frac{\ln\left(1+x\right)\ln\left(1+\frac{1}{x^2}\right)}{x}\,dx=\int_0^{1}\frac{\ln\left(\frac{1+x}{x}\right)\ln\left(1+{x^2}\right)}{x}\,dx$$ $$I=2\int_0^{1}\frac{\ln\left(1+x\right)\ln\left(1+{x^2}\right)}{x}\,dx-\frac{9}{4}\int_0^{1}\frac{\ln\left(1+x\right)\ln\left(x\right)}{x}\,dx={\pi}G-\frac{3}{8}\zeta(3)$$ These integrals are known $$\int_0^{1}\frac{\ln\left(1+x\right)\ln\left(1+{x^2}\right)}{x}\,dx=-\frac{33}{32}\zeta(3)+\frac{\pi}{2}G,\int_0^{1}\frac{\ln\left(1+x\right)\ln\left(x\right)}{x}\,dx=-\frac{3}{4}\zeta(3)$$

Ali Shadhar
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