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Are we aware of an elementary way of proving that?

$$\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$$

Of course, with the help of Mathematica it can be done, but I wonder if there exists an elementary, simple, easy way of finishing it.

$$\int \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=$$ $$=\text{Li}_3\left(\frac{t+2}{t}\right)-\text{Li}_3\left(-\frac{t+2}{t}\right)+\text{Li}_3(-t)+\text{Li}_3(t+2)-\left(\text{Li}_2\left(\frac{t+2}{t}\right)-\text{Li}_2\left(-\frac{t+2}{t}\right)\right) \log \left(-\frac{t+2}{t}\right)-\text{Li}_2(-t-1) \left(\log \left(\frac{1}{t}\right)+\log (t)\right)-\text{Li}_2(-t) \left(\log (t+2)-\log \left(-\frac{t+2}{t}\right)\right)-\text{Li}_2(t+2) \left(\log \left(-\frac{t+2}{t}\right)+\log (t)\right)-\frac{1}{2} \left(-\log \left(\frac{2 (t+1)}{t}\right)+\log \left(-\frac{2}{t}\right)+\log (t+1)\right) \log ^2\left(-\frac{t+2}{t}\right)-(\log (-t-1)-\log (t+1)) \log (t+2) \log \left(-\frac{t+2}{t}\right)-\log (t) \log (t+1) \log (t+2)-\frac{1}{2} (\log (t+1)-\log (-t-1)) \log (t+2) (\log (t+2)-2 \log (t))$$

that is obtain by Mathematica.

EDIT: I think once we rewrite all as

$$ \frac{1}{2}\int_0^1 \left(\frac{\log ^2\left(\frac{t+2}{t}\right)}{t+1}-\frac{\log ^2(t+2)}{t+1}-\frac{\log ^2\left(\frac{1}{t}\right)}{t+1}\right) \, dt$$ and employ the proper variable change, we're almost done, the idea that just crossed my mind.

The hardest part is probably the integral $$\int_0^1 \frac{\log ^2\left(\frac{t+2}{t}\right)}{t+1} \ dt$$ where by letting $t/(t+2)\mapsto t$, we obtain the far nicer form

$$2\int_0^{1/3} \frac{\log ^2(t)}{1-t^2} \, dt$$ $$=\int_0^{1/3} \frac{\log ^2(t)}{ 1-t} \ dt+\int_0^{1/3}\frac{\log ^2(t)}{ 1+t} \, dt$$ where then using geometric series and swapping the integration and summation, we obtain that$$\int_0^1 \frac{\log ^2\left(\frac{t+2}{t}\right)}{t+1} \ dt=$$ $$2\text{Li}_3\left(\frac{1}{3}\right)-2\text{Li}_3\left(-\frac{1}{3}\right)+ \log (3) \left(4 \text{Li}_2\left(\frac{1}{3}\right)-\text{Li}_2\left(\frac{1}{9}\right)+\log (2) \log (3)\right) \tag1$$ Using geometric series again, we easily get that

$$\int_0^1 \frac{\log ^2\left(\frac{1}{t}\right)}{t+1} \, dt=\frac{3}{2} \zeta (3) \tag2$$ The last integral can be reduced by variable change to a well-known integral treated by Leonard Lewin in his book about polylogarithms, that is $$\int_0^1 \frac{\log ^2(t+2)}{t+1} \, dt=\int_1^2 \frac{\log ^2(t+1)}{t} \, dt$$ So, we can use the fact that $$\int_0^t \frac{\log ^2(1+t)}{t} \, dt$$ $$=\log(t)\log^2(1+t)-2/3\log^3(1+t)-2\log(1+t)\operatorname{Li}_2\left(\frac{1}{1+t}\right)-2\operatorname{Li}_3\left(\frac{1}{1+t}\right)+2\zeta(3)$$

So, using the foregoing result, we immediately get that $$\int_1^2 \frac{\log ^2(t+1)}{t} \, dt$$ $$=-2 \text{Li}_3\left(\frac{1}{3}\right)-2 \text{Li}_2\left(\frac{1}{3}\right) \log (3)+\frac{7 \zeta (3)}{4}-\frac{1}{3} 2 \log ^3(3)+\log (2) \log ^2(3) \tag3$$

Combining $(1) (2) (3)$, we obtain that $$\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt$$ $$=2 \text{Li}_3\left(\frac{1}{3}\right)-\text{Li}_3\left(-\frac{1}{3}\right)-\frac{1}{2} \left(\text{Li}_2\left(\frac{1}{9}\right)-6 \text{Li}_2\left(\frac{1}{3}\right)\right) \log (3)-\frac{13 \zeta (3)}{8}+\frac{\log ^3(3)}{3}$$

Using $(20)$ from here http://mathworld.wolfram.com/Dilogarithm.html and then the identity and $(1)$ from this question/answer Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$?, we get that

$$\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3).$$

A SPECIAL ADDENDUM: The present integral can be viewed as the key core of calculating a pretty tough integral that was posted on MSE some time go, $$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ at this link Evaluating $\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$.

Simple integration by parts combined with the use of the geometric series shows that

$$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ $$=\int_0^1 \frac{\log (z+2) \log (z)}{z+1}dz+\underbrace{\int_0^1\frac{\log (2-z) \log (z)}{1-z} \ dz}_{\large \sum _{n=1}^{\infty } \frac{(-1)^n H_n}{n^2}=-5/8 \zeta (3)}-\underbrace{\int_0^1\frac{\log (1-z) \log (z)}{1-z}dz}_{\large\sum _{n=1}^{\infty } \frac{H_n}{(n+1)^2}=\zeta (3)}-\underbrace{\int_0^1\frac{\log (z+1) \log (z)}{z+1} \, dz}_{\large \sum _{n=1}^{\infty } \frac{(-1)^n H_n}{(n+1)^2}=-1/8\zeta (3)}$$ where since the integral unknown is calculated in this post, we conclude that

$$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ $$=-\frac{49}{24}\zeta(3).$$ A SPECIAL ADDENDUM $2$: By a careful approach of the previous results, we also obtain the value of the very beautiful series

$$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^k }{k2^k n 2^n (k+n)}=\frac{1}{12} \left(3 \text{Li}_3\left(\frac{1}{4}\right)+6\log (2) \text{Li}_2\left(\frac{1}{4}\right) +4 \log ^3(2)-4 \zeta (3)\right)$$

Q.E.D.

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  • $\begingroup$ @AlecTeal all that crossed my mind, but nothing is useful. I tried to turned it into a series, say, but that didn't seem useful at all. $\endgroup$ – user 1357113 Jun 30 '15 at 12:06
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    $\begingroup$ It looks like you are almost done, you just need to find a way to use the dilogarothm/trilogarithm functional identities to get rid of every $\text{Li}_2$ or $\text{Li}_3$. $\endgroup$ – Jack D'Aurizio Jun 30 '15 at 13:37
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    $\begingroup$ @JackD'Aurizio Yes, you're right! $\endgroup$ – user 1357113 Jun 30 '15 at 13:39
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    $\begingroup$ Using the substitution $x\mapsto x^{-1}(1-x)$, you could just write the integral as $$\frac{1}{2}\int^1_\frac{1}{2}\frac{\ln^2(1-x)+\ln^2(1+x)+2\ln^2{x}-2\ln{x}\ln(1-x^2)-\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx$$ and all of these resultant integrals are relatively straightforward to compute. $\endgroup$ – M.N.C.E. Jul 1 '15 at 10:36
  • $\begingroup$ Aware...aware... like Jean-Claude Van Damme? :P $\endgroup$ – Hexacoordinate-C Nov 4 '15 at 19:38
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The last series Special Addendum in the OP looks quite interesting! (just an observation, not an answer)

We have by symmetry of the series:

$$S = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^k }{k2^k n 2^n (k+n)} = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^n}{k2^k n 2^n (k+n)}$$

Hence, $$\begin{align}\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{1+(-1)^k+(-1)^n+(-1)^{k+n} }{k2^k n 2^n (k+n)} &= \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(1+(-1)^k)(1+(-1)^n)}{k2^k n 2^n (k+n)}\\&= \frac{1}{2}\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{1}{k4^k n 4^n (k+n)}\end{align}$$

Since, the term $(1+(-1)^k)(1+(-1)^n)$ survives only when both $k,n$ are even.

Since, $$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{a^{n+k}}{kn(k+n)} = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \int_0^1 \frac{a^{n+k}x^{n+k}}{kn}\frac{dx}{x} = \int_0^1 \frac{\log^2 (1-ax)}{x}\,dx$$

We have: $$2S = -\int_0^1 \frac{\log^2(1-\frac{x}{2})}{x}\,dx-\int_0^1 \frac{\log^2(1+\frac{x}{2})}{x}\,dx+\frac{1}{2}\int_0^1 \frac{\log^2(1-\frac{x}{4})}{x}\,dx$$

The integrals for $0 < a < 1$:

$\displaystyle \begin{align} \int_0^1 \frac{\log^2 (1-ax)}{x}\,dx &= \int_0^a \frac{\log^2 (1-x)}{x}\,dx \\&= \log a \log^2(1-a)+2\int_0^a \frac{\log x \log(1-x)}{1-x}\,dx\\&= \log a \log^2(1-a)+2\log (1-a) \operatorname{Li}_2(1-a) - 2\operatorname{Li}_3(1-a)+2\operatorname{Li}_3(1)\end{align}$

and the integrals for $0 < a < 1$:

$\displaystyle \begin{align} \int_0^1 \frac{\log^2 (1+ax)}{x}\,dx &= \int_0^{a} \frac{\log^2 (1+x)}{x}\,dx \\&= \log a \log^2(1+a)-2\int_0^a \frac{\log x \log(1+x)}{1+x}\,dx\\&= \log a \log^2 (1+a) - \frac{2}{3}\log^3 (1+a)-2\log (1+a)\operatorname{Li}_2\left(\frac{1}{1+a}\right)-2\operatorname{Li}_3\left(\frac{1}{1+a}\right)+2\operatorname{Li}_3(1)\end{align}$

Wolfram alpha computes these three integrals as:

[1] $$\displaystyle \int_0^1 \log^2 \left(1+\frac{x}{2}\right)\,\frac{dx}{x}=-2 \text{Li}_3\left(\frac{2}{3}\right)-\text{Li}_2\left(\frac{2}{3}\right) \log \left(\frac{9}{4}\right)+2 \zeta (3)\\+\frac{4}{3} \log (18) \coth ^{-1}(5)^2$$

[2] $$\int_0^1 \log^2\left(1-\frac{x}{2}\right)\,\frac{dx}{x} = \frac{1}{4}\zeta(3) - \frac{1}{3}\log^3 (2)$$

[3] $$\int_0^1 \log^2\left(1-\frac{x}{4}\right)\,\frac{dx}{x} = -2 \text{Li}_3\left(\frac{3}{4}\right)-\text{Li}_2\left(\frac{3}{4}\right) \log \left(\frac{16}{9}\right)+2 \zeta (3)\\-\log^2(3)\log (4)+\log^2(4)\log\left(\frac{9}{4}\right)$$

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    $\begingroup$ Good to know different approaches! (+1) :-) $\endgroup$ – user 1357113 Jun 30 '15 at 15:56
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I used my answer Products of logarithms under the integral sign where I set $(A,B,C)=(-1,0,2)$ and the limits to $(x_1,x_2)=(\epsilon,1-\epsilon)$. This generated a bunch of terms involving logarithms and poly-logarithms . I expanded those terms and carefuly took the limit $\epsilon \rightarrow 0$. After that i applied the inversion formula for poly-logarithms http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/17/02/01/01/0005/ to reduce the argument of all polylogs to the domain $(-1,1)$ and after that I simplified the result in Mathematica. what came out was the following: \begin{eqnarray} &&\int\limits_0^1 \frac{\log(x) \log(x+2)}{x+1} dx = \\ && \text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)+\left(\text{Li}_2\left(-\frac{1}{3}\right)-2 \text{Li}_2\left(\frac{1}{3}\right)\right) \log (3)+\frac{13 \zeta (3)}{8}-\frac{1}{3} \log ^3(3) \end{eqnarray} Now we have: \begin{eqnarray} \text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right) &=& - \frac{\log(3)^3}{6} + \frac{\pi^2}{6} \log(3) - \frac{13}{6} \zeta(3) \quad (i)\\ \text{Li}_2\left(-\frac{1}{3}\right)-2 \text{Li}_2\left(\frac{1}{3}\right) &=& \frac{\log(3)^2}{2} - \frac{\pi^2}{6} \quad (ii) \end{eqnarray} where in $(i)$ we used Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$? and in $(ii)$ we used the Abel identity https://en.wikipedia.org/wiki/Polylogarithm for $(x,y)=(-1,-2)$ along with the reflection and the inversion formulae for the dilogarithm. Combining everything together we get: \begin{equation} \int\limits_0^1 \frac{\log(x) \log(x+2)}{x+1} dx = - \frac{13}{24} \zeta(3) \end{equation} as it should be.

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  • $\begingroup$ Thanks for the answer (I noticed this answer with much delay). $\endgroup$ – user 1357113 May 24 at 10:48
  • $\begingroup$ @user 1357113 Out of curiosity how did you know that this very integral has such a simple closed form. I have been playing around with integrals of that kind by replacing the one and the two by some other integers and using the anti-derivative in the first link top above but i never get something as simple as you have; i.e. there is always a bunch of poly-logs which are hard to eliminate. So again how did you know that this very specimen has such a simple closed form ? $\endgroup$ – Przemo May 24 at 12:02
  • $\begingroup$ @Przemo: Do you know what is LLL algorithm? $\endgroup$ – FDP May 24 at 16:28
  • $\begingroup$ @Przemo one of my strongest traits is the intuition that works in a way hard to explain. I do not use any LLL algorithm (mentioned by @FDP). This is only a small such result, but out there are many results of this kind. $\endgroup$ – user 1357113 May 27 at 17:53
  • $\begingroup$ LLL allows to search for such integrals wich provide nice results. But sometimes when you try to compute something you find out some nice results:\begin{align}\int_0^{+\infty} \frac{\arctan x\ln x}{1+x^2}\,dx=\frac{7}{8}\zeta(3)\end{align} $\endgroup$ – FDP May 27 at 20:17

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