<s>[INST] Evaluate $\int_1^{\infty} {\frac{\ln{x}}{(x-1)(2x-1)}\,dx}$ Evaluate:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}$$
It turns out that the value of the integral is exactly:
$$\frac{1}{12}\left(\pi^2+6\ln^2{2}\right)$$
as found by WolframAlpha, but Wolfram gives no indication of how it arrives at this curious result.
How would one solve the integral analytically? A first step might be the $u$-sub, $u=\frac{1}{x}$, which gives:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}=-\int_0^1 {\left(\frac{\ln{x}}{\left(1-x\right)\left(2-x\right)}\,dx\right)}$$
Thanks!
 [/INST] This is related to the dilogarithm function $\text{Li}_2(x)$ and the special value $$\text{Li}_2(\frac{1}{2}) = \frac{\pi^2}{12}-\frac{\ln^2 2}{2}$$
The dilogarithm has integral representation:
$$\text{Li}_2(z) = \frac{\pi^2}{6}+\int_0^{1-z} \frac{\ln t}{1-t} dt$$
For your question, we have
$$\begin{aligned}
\int_0^1 \frac{\ln{x}}{\left(1-x\right)\left(2-x\right)}dx &= \int_0^1 \frac{\ln x}{1-x}dx - \int_0^1\frac{\ln x}{2-x}dx \\
&= \int_0^1 \frac{\ln x}{1-x}dx - 2\int_0^{1/2}\frac{\ln 2x}{2-2x}dx \\
&= \int_0^1 \frac{\ln x}{1-x}dx-(\ln 2) \int_0^{1/2}\frac{1}{1-x}dx-\int_0^{1/2}\frac{\ln x}{1-x}dx \\
&= -\frac{\pi^2}{6}-\ln^2 2 -[\text{Li}_2(\frac{1}{2})-\frac{\pi^2}{6}] \\
&= -\frac{\pi^2}{12}-\frac{\ln^2 2}{2}
\end{aligned}$$
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