How to acomplish the evaluation of $$\int _0^{\infty }\frac{\ln \left(1+x^3\right)}{x\left(1+x^2\right)}\:dx$$ I got the integral to be $$\int _0^{\infty }\frac{\ln \left(1+x^3\right)}{x}\:dx-\int _0^{\infty }\frac{\ln \left(1+x^3\right)}{1+x^2}\:dx$$ But im stuck, can someone give me a hint on how to proceed?
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$\begingroup$ It's not valid to split the integral up like this in this case, since ${\int_{0}^{\infty}\frac{\ln(1+x^3)}{x}dx}$ diverges. We can only split it up like this if both individual integrals converge, which is not true in this case. I don't think any solution method will be particularly "elementary" (depending on what you define as elementary) in this problem $\endgroup$ – Riemann'sPointyNose Aug 17 '20 at 1:59
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$\begingroup$ Have you tried the substitution $x = \tan \theta$? $\endgroup$ – John White Aug 17 '20 at 2:13
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$\begingroup$ The integral in this post looks very similar. $\endgroup$ – Viktor Vaughn Aug 17 '20 at 5:30
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Note
$$ \int _1^{\infty }\frac{\ln \left(1+x^3\right)}{x\left(1+x^2\right)}\:dx \overset{x\to\frac1x} = \int _0^{1}\frac{x\ln \left(1+x^3\right)}{1+x^2}\:dx - \int _0^{1}\frac{3x\ln x}{1+x^2}\:dx $$
Then
\begin{align} \int _0^{\infty }\frac{\ln \left(1+x^3\right)}{x\left(1+x^2\right)}\:dx & = \int _0^{1}\frac{\ln \left(1+x^3\right)}{x}\:dx - \int _0^{1}\frac{3x\ln x}{1+x^2}\:dx \\ & = \int _0^{1}\frac{\ln \left(1+x^3\right)}{x}\:dx +\frac32 \int _0^{1}\frac{\ln (1+x^2)}{x}\:dx \\ & = \frac13 \int _0^{1}\frac{\ln \left(1+t\right)}{t}\:dt +\frac34\int _0^{1}\frac{\ln (1+t)}{t}\:dt \\ & = \frac{13}{12} \int _0^{1}\frac{\ln (1+t)}{t}\:dt \\ &= \frac{13}{12}\cdot \frac{\pi^2}{12}= \frac{13\pi^2}{144} \end{align}
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$\begingroup$ It works nice using Feynman's trick. Cheers :-) $\endgroup$ – Claude Leibovici Aug 17 '20 at 10:21
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Using Feynman's trick.
$$I(a)=\int _0^{\infty }\frac{\log \left(1+a^3x^3\right)}{x\left(1+x^2\right)}\:dx$$ $$I'(a)=\int _0^{\infty }\frac{3a^2x^2}{\left(1+x^2\right) \left(1+a x^3\right)}\,dx$$ Use partial fraction decomposition, the integrand write $$\frac{a^2}{\left(a^2+1\right) (a x+1)}-\frac{3 \left(a^5 x+a^2\right)}{\left(a^2+1\right) \left(a^4-a^2+1\right) \left(x^2+1\right)}+\frac{(2 a^5-a^3) x-a^4+2 a^2}{\left(a^4-a^2+1\right) \left(a^2 x^2-a x+1\right)}$$ Computing $I'(a)$ is quite simple and the final result is $$I'(a)=\frac{2 \pi a}{\sqrt{3} \left(a^6+1\right)}+\frac{3 a^5 \log (a)}{a^6+1}+\frac{2 \pi a^3}{\sqrt{3} \left(a^6+1\right)}-\frac{3 \pi a^2}{2 \left(a^6+1\right)}$$
For sure, the integration with repect to $a$ is not the most pleasant but everything simplifies at the bounds $$J=\int_0^1 I'(a)\,da=\Big[\frac{1}{144} \left(36 \left(\text{Li}_2\left((-1)^{1/3}\right)+\text{Li}_2\left(-(-1)^{2/3}\right) \right)-53 \pi ^2\right) \Big] -\Big[-\frac{4 \pi ^2}{9} \Big]$$ $$J=-\frac{17 \pi ^2}{48}+\frac{4 \pi ^2}{9}=\frac{13 \pi ^2}{144}$$
answered Aug 17 '20 at 11:02
