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I was trying to evaluate the definite integral

$$\int_0^{\frac{1}{2}} \frac{1}{x} \cdot \ln(1+x) \cdot \ln\left(\frac{1}{x} -1\right)\,\mathrm{d}x$$

On WolframAlpha, I found out that this converges to 0.651114

Which seems to be pretty close (and possibly equal) to $\frac{13}{24}\cdot\zeta(3)$

Where $\zeta(3)$ is the value of the Riemann Zeta function $\zeta(n)$ at $\ n = 3$

Is this true? How can we prove that they are equal?

I tried substituting $\ x = e^{-t}$ and the integral became

$$\ I = \int_0^{\ln2}\ln(1+e^{-t}) \cdot \ln\left(\frac{e^{-t}}{1-e^{-t}}\right)\,\mathrm{d}t$$

How do I proceed further?

Any help is appreciated. Thanks for reading.

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    For information, Mathematica gives a closed form solution for the primitive of the integrand: Log[-1 + x] Log[x] Log[1 + x] + 1/2 (-Log[-x] + Log[x]) Log[ 1 + x] (-2 Log[-1 + x] + Log[1 + x]) + (Log[-x] - Log[x]) Log[ 1 + x] Log[-((1 + x)/(-1 + x))] + 1/2 (Log[-(2/(-1 + x))] + Log[x] - Log[(2 x)/(-1 + x)]) Log[-(( 1 + x)/(-1 + x))]^2 + (Log[1 + x] - Log[-((1 + x)/(-1 + x))]) PolyLog[2, 1 - x] + Log[x] PolyLog[2, -x] - .... – NN2 Dec 16 '23 at 18:33
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    ... (-Log[-1 + x] + Log[(1 - x)/x] + Log[x]) PolyLog[ 2, -x] + (Log[-1 + x] + Log[-((1 + x)/(-1 + x))]) PolyLog[2, 1 + x] + Log[-((1 + x)/(-1 + x))] (-PolyLog[2, -((1 + x)/(-1 + x))] + PolyLog[2, (1 + x)/(-1 + x)]) - PolyLog[3, 1 - x] - PolyLog[3, -x] - PolyLog[3, 1 + x] + PolyLog[3, -((1 + x)/(-1 + x))] - PolyLog[3, (1 + x)/(-1 + x)] – NN2 Dec 16 '23 at 18:33
  • @NN2 That looks very intimidating. –  Dec 16 '23 at 20:33

3 Answers3

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I'm outsourcing most of the work from another answer. Perform integration by parts with $f = \ln(\frac{1}{x}-1) =\ln(1-x)-\ln(x) $ and $g' = \frac{\ln(1+x)}{x}$ to observe

$$\begin{align}\int_0^{1/2} \frac{1}{x}\ln(1+x)\ln\left(\frac{1}{x}-1\right)dx &= -\ln\left(\frac{1}{x}-1\right)\text{Li}_2(-x)\bigg\vert_{0}^{1/2} -\int_0^{1/2}\frac{\text{Li}_2(-x)}{x}+\frac{\text{Li}_2(-x)}{1-x} dx\\ &=-\int_0^{1/2}\frac{\text{Li}_2(-x)}{x}dx-\int_0^{1/2}\frac{\text{Li}_2(-x)}{1-x} dx\\ \end{align}$$

We have $$-\int_0^{1/2}\frac{\text{Li}_2(-x)}{x}dx =-\text{Li}_3\left(-\frac{1}{2}\right)$$ more or less by definition and the second is seen here to be given by

$$-\int_0^{1/2}\frac{\text{Li}_2(-x)}{1-x}dx = \text{Li}_3\left(-\frac{1}{2}\right) + \frac{13}{24}\zeta(3) $$

Edit: Looking at Bob's answer, it is really the calculation of his integral that is most of the work in the separate question I linked.

podiki
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    Thank you for the answer. What a ride it has been, the accepted answer to the linked question itself linked outsourced its calculation to another answer of another question. –  Dec 16 '23 at 22:05
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This integral is not so bad if we use Feynman's trick.

$$I(a)=\int_{0}^{\frac 12} \frac{1}{x} \, \ln(1+ax) \, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\int_{0}^{\frac 12} \frac 1 {1+ax}\, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}$$ $$I(a)=\int_0^1 \frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}\,da$$ The corresponding antiderivative is not bad (in terms of logarithms and polylogarithms) and $$\color{blue}{I(1)=\frac{\log ^3(3)}{3}-\frac{13}{8} \zeta (3)+\left(2 \text{Li}_2\left(\frac{1}{3}\right)-\text{Li}_2\left(-\frac{1} {3}\right)\right) \log (3)+}$$ $$\color{blue}{\left(2 \text{Li}_3\left(\frac{1}{3}\right)-\text{Li}_3\left(-\frac{1} {3}\right)\right)}$$

Edit

I do not know how to reduce the result to $\frac{13 }{24}\zeta (3)$ which is exact for at least $10,000$ significant figures.

Claude Leibovici
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    Thank you for your answer, Claude Leibovici. Where can I learn more about this polylogarithm function? –  Dec 18 '23 at 07:27
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Not an answer... Too long for a comment.

We can split the given integral as $I_1-I_2.$

The second piece is easy. By integration by parts, we have

$$\int\frac{\ln(1+x)\ln x}x dx\\=-\text{Li}_2(-x)\ln x+\int\frac{\text{Li}_2(-x)}xdx\\=-\text{Li}_2(-x)\ln x+\text{Li}_3(-x)+C$$ as in WolframAlpha's output. Hence, $I_2=\text{Li}_2(-\frac12)\ln 2+\text{Li}_3(-\frac12)\approx-0.783415.$

For the first piece WA gave a complicated antiderivative output and I questioned the purpose of my life. Why should İ try to find it? Anyway, we can do series expansion $$\int_0^{\frac12}\frac{\ln(1+x)\ln(1-x)}x dx\\=-\sum_{n=1}^\infty \frac1{n}\int_0^{\frac12}x^{n-1}\ln(1+x)dx$$ The integrals in the sum are doable by integration by parts for numerical approximation. WA gives $I_1\approx-0.132301$.

Unfortunately, I cannot see any connection with the value of Riemann Zeta Function at 3.

Bob Dobbs
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