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I am trying to prove this interesting integral $$ I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}. $$ I tried using $y=1+x^3$ but that didn't help. We can possibly try $$ I=\int_0^\infty \frac{\log(1+x^3) x}{1+x^3} \,dx-\int_0^\infty \frac{\log(x^3) x}{1+x^3}\,dx. $$ These integrals would be much easier had the bounds been from $0 $\ to $\infty$, however they are not. Perhaps partial integration will work but I didn't find the way if we try $$ dv=\frac{x}{1+x^3}, \quad u= \log(1+x^3) $$ but I ran into a divergent integral. Thanks how can we prove I?

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  • $\begingroup$ Given that the argument of the log is $\frac{1+x^3}{x^3}=1+x^{-3}$, did you try a u-substitution $u=\frac1x, du=-\frac{1}{x^2}dx$? $\endgroup$ – Steven Stadnicki May 20 '14 at 21:00
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Let us make the change of variables $$v=\frac{x^3}{1+x^3}\iff x=\left(\frac{v}{1-v}\right)^{1/3}$$ This transforms the integral $I$ to the following form $$ I=-\frac{1}{3}\int_0^1\log(v)\,v^{-1/3}(1-v)^{-2/3}dv $$ Now, If $$f(\alpha):=B(\alpha,\frac{1}{3})=\int_0^1v^{\alpha-1}(1-v)^{\frac{1}{3}-1}dv=\frac{\Gamma(\alpha)\Gamma(\frac{1}{3})}{\Gamma(\alpha+\frac{1}{3})}$$ then $I=-\frac{1}{3}f'(\frac{2}{3})$. But, since $\Gamma(1)=1$, and $\Gamma'(1)=-\gamma$, we have $$\eqalign{ f'\left(\frac{2}{3}\right)&=\Gamma'\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)-\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)\Gamma'(1)\cr &=\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)\left(\psi\left(\frac{2}{3}\right)+\gamma\right)\cr &\buildrel{\rm(1)}\over{=}\frac{\pi}{\sin(\pi/3)}\left(\psi\left(\frac{2}{3}\right)+\gamma\right)\cr &\buildrel{\rm(2)}\over{=}\frac{2\pi}{\sqrt{3}}\left( \frac{\pi}{2\sqrt{3}}-\frac{3}{2}\log 3\right) } $$ Where, for $(1)$ we used the Euler's reflection formula, and for $(2)$ we used Gauss' theorem for the digamma function. Combining our results we get $$ I=-\frac{1}{3}\frac{2\pi}{\sqrt{3}}\left( \frac{\pi}{2\sqrt{3}}-\frac{3}{2}\log 3\right)=\frac{\pi}{\sqrt{3}}\log 3-\frac{\pi^2}{9}. $$ which is the desired result.$\qquad\square$

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  • $\begingroup$ Beautiful solution +1 Thank you for this $\endgroup$ – Jeff Faraci May 20 '14 at 22:54
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Define $$ I(a)=\int_0^\infty \log \frac{a+x^3}{x^3} \frac{x \,dx}{1+x^3}.$$ Then $I(0)=0$ and \begin{eqnarray} I'(a)&=&\int_0^\infty\frac{x}{(a+x^3)(1+x^3)}dx\\ &=&\frac13\int_0^\infty\frac{1}{x^{1/3}(a+x)(1+x)}dx\\ &=&\frac{1}{3(1-a)}\left(\int_0^\infty\frac{1}{x^{1/3}(a+x)}dx-\int_0^\infty\frac{1}{x^{1/3}(1+x)}dx\right)\\ &=&\frac{1}{3(1-a)}\frac{2\pi}{\sqrt3}\left(\frac{1}{a^{1/3}}-1\right)\\ &=&\frac{2\pi}{3\sqrt3}\frac{1}{a+a^{2/3}+a^{1/3}} \end{eqnarray} Here we use $$ \int_0^\infty\frac{1}{x^{1/3}(a+x)}dx=\frac{2\pi}{\sqrt3 a^{1/3}}. $$ Thus \begin{eqnarray} I(1)&=&\frac{2\pi}{\sqrt3}\int_0^1 \frac{1}{a+a^{2/3}+a^{1/3}}da\\ &=&\frac{2\pi}{3\sqrt3}\int_0^1 \frac{b}{b^2+b+1}db\\ &=&-\frac{\pi^2}{9}+\frac{\pi}{\sqrt3}\ln 3. \end{eqnarray}

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  • $\begingroup$ Very clever way (+1) $\endgroup$ – Venus Dec 11 '14 at 5:40

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