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.
Best Answer
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.