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I was trying to obtain some interesting identity and refresh my mathematics, as same manner and tricks showed in [1] for modified functions (my modified function that I write in the integrand and my modified Fact, see My attempt if you want/need to know it). My question is

Question. Can you to evaluate $$\int_0^1\left(\int_0^\infty\frac{x^2}{(x^3z^3+1)(x^3+1)}dx\right)dz.$$ Thanks in advance.

My attempt: First we prove

Fact. The algebraic identity holds $$\frac{1}{z^3-1}\left(\frac{2x^2z^3}{x^3z^3+1}-\frac{2x^2}{x^3+1}\right)=\frac{2x^2(z^3-1)}{(x^3z^3+1)(x^3+1)}.$$

Thus previous integral reduces to $$\int_{0}^1\frac{\log z}{z^3-1}dz.$$ By this way I can use/work a change of variable $u=\log z$ and $dv=dz/(z^3-1)$, decomposition in simple fraction to $1/(z^3-1)=a/(z-1)+B/(z^2+z+1)$, gives $A=1/3,B=-2/3$ and it is known that $\int dx/(1+x+x^2)=2\arctan((2x+1)/\sqrt{3})/\sqrt{3}$. But I obtain troubles to evaluate the integration limits.

In the other hand, using Fubini we have $$\int_0^\infty\frac{x^2}{x^3+1}\left(\int_0^1\frac{dz}{x^3z^3+1}\right)dx,$$ then I can work the change $u^2=x^3z^3$, $dz=2zdu/3u$ to write $$\int_0^1\frac{dz}{x^3z^3+1}=\frac{1}{x}\int_{0}^{x^{3/2}}\frac{2}{3}\frac{u^{-1/3}}{u^2+1}du.$$

Summarizing I don't how continue to evaluate at least one time the integral (the author of [1] uses two times Fubini and a MacLaurin expansion to obtain his statement). I do not think that my modification will be useful, but I want to learn to integrate at least to one manner previous integral (if it is possible find an identity is the best (is not required))

References:

[1] James Harper, Another Simple Proof of $1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{6}$, The American Mathematical MONTHLY, 2003, pages 540-541.

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  • $\begingroup$ Please, do not close or delete the question. It is a nice problem which could be of a lot of interest for many participants. By the way, I am next door to you ! $\endgroup$ – Claude Leibovici Nov 2 '15 at 11:38
  • $\begingroup$ My english is bad, to me close is choose an answer @ClaudeLeibovici $\endgroup$ – user243301 Nov 2 '15 at 11:41
  • $\begingroup$ Mejor, amigo ! Hasta pronto. I honestly think that Achille's answer is better than mine. $\endgroup$ – Claude Leibovici Nov 2 '15 at 11:48
  • $\begingroup$ @achillehui can you tell me explicitely what have I to prove to use Tonelli in your answer? Then I can save my computations in a comment and don't disturb to you more, thanks. I am trying use Tonelli in other recent post, I know that when I was student I studied it, and I put my eyes on en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli.27s_theorem but I want know in easy words what statements I have to prove. Very thanks much $\endgroup$ – user243301 Nov 3 '15 at 19:36
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Substitute $z = e^{-t}$,

$$\begin{align} \int_0^1 \frac{\log z}{z^3-1} dz &= \int_0^\infty t \frac{e^{-t}}{1-e^{-3t}}dt =\int_0^\infty t \left(\sum_{k=0}^\infty e^{-(3k+1)t}\right) dt\\ \color{blue}{\text{by Tonelli}\;\rightarrow}\quad &= \sum_{k=0}^\infty \int_0^\infty te ^{-(3k+1)t} dt\\ &= \sum_{k=0}^\infty \frac{1}{(3k+1)^2} = \frac19 \sum_{k=0}^\infty \frac{1}{(k+\frac13)^2}\\ &= \frac19 \zeta\left(2,\frac13\right) \end{align} $$ where $\zeta( s, q ) = \sum_{n=0}^\infty \frac{1}{(n+q)^s}$ is the Hurwitz Zeta function.

Throw the expression 1/9*HurwitzZeta[2,1/3] to WA, the integral $$\approx 1.121733013936343786865778233321390706724322679920108682437964$$

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  • $\begingroup$ Very thanks much @achillehui I take notes now and in some hours I will close the question. $\endgroup$ – user243301 Nov 2 '15 at 11:22
  • $\begingroup$ This is very nice approach ! Thanks for providing it. $\endgroup$ – Claude Leibovici Nov 2 '15 at 11:37
  • $\begingroup$ I choose achille's answer, honestly I understand it better, but I hope in future times to try put on game your method. Thanks @ClaudeLeibovici como venía a decir Seneca en Cartas a Lucilio, el gladiador no hace estrategias antes de ir a la arena, espera a ver qué condiciones y armas tiene su oponente, incluso la caída de cuerpo de éste, solo entonces actúa. En resumen habéis demostrado ser unos gladiadores de las matemáticas porque....los gladiadores no hacen futurizos. $\endgroup$ – user243301 Nov 2 '15 at 14:15
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Conserning $$I=\int\frac{\log z}{z^3-1}dz$$ I think that the problem becomes simpler if we are allowed to use the fact that $$\int\frac{\log z}{z-a_i}dz=\text{Li}_2\left(\frac{z}{a_i}\right)+\log (z) \log \left(1-\frac{z}{a_i}\right)$$ which makes $$\int_0^1\frac{\log z}{z-a_i}dz=\text{Li}_2\left(\frac{1}{a}\right)$$ provided that $\Im(a_i)\neq 0\lor \Re(a_i)\geq 1\lor \Re(a_i)\leq 0$, using for the $a_i$'s $$a_1= 1,\quad a_2= -(-1)^{1/3},\quad a_3= (-1)^{2/3}$$ So, partial fraction decomposition and, by the end, this makes $$I=\int_0^1\frac{\log z}{z^3-1}dz=\frac{4 \pi ^2}{81}+\frac{2 \psi ^{(1)}\left(\frac{1}{3}\right)}{27}-\frac{\psi ^{(1)}\left(\frac{2}{3}\right)}{27}$$

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  • $\begingroup$ Very thanks much too to you @ClaudeLeibovici, I invite to you and previous user if you want discuss some detail, since you are professionals, I hace that real and write notes on your answer. $\endgroup$ – user243301 Nov 2 '15 at 11:26

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