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Compute $$I = \int_1^\infty \log^2 \left(1-\frac 1 x\right) \, dx$$ I made the substitution: $$t=\frac 1 x$$ It follows: $$I=\int_0^1 \frac{\log^2(1-t)}{t^2} \, dt$$ My next step would be to compute the derivative of the following integral with parameter $y$, w.r.t to $y$: $$F(y)=\int_0^1 \frac{\log^2(y-t)}{t^2} \, dt$$ Or something like this. I think would be a nice solution to use this kind of approach. But I am getting stuck after computing the derivative.

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Once you get $$ I = \int_{0}^{1}\frac{\log^2(1-t)}{t^2}\,dt=\int_{0}^{1}\frac{\log^2(t)}{(1-t)^2}\,dt \tag{1}$$ it is enough to recall that $$ \frac{1}{(1-t)^2}=\sum_{n\geq 0}(n+1) t^n,\qquad \int_{0}^{1}t^n\log^2(t)\,dt = \frac{2}{(n+1)^3}\tag{2} $$ to have: $$ I = \sum_{n\geq 0}\frac{2}{(n+1)^2} = \color{red}{\frac{\pi^2}{3}}.\tag{3}$$

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    $\begingroup$ That is the answer! $\endgroup$ – penguina Jan 22 '17 at 23:23
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I thought it might be instructive to present an approach that relies on integration by parts and recognition of the value of the integral, $\int_0^1 \frac{\log(1-x)}{x}\,dx=\text{Li}_2(1)=\frac{\pi^2}{6}$. To that end, we proceed.


Let $I$ be given by

$$I=\int_1^\infty \log^2\left(1-\frac1x\right)\,dx=\int_0^1\frac{\log^2(1-x)}{x^2}\,dx$$

Integrating by parts with $u=\log^2(1-x)$ and $v=-1/x$ reveals

$$\begin{align} I&=\lim_{\epsilon\to 0}\left(\left.\left(-\frac{\log^2(1-x)}{x}\right)\right|_{0}^{1-\epsilon}-2\int_0^{1-\epsilon}\frac{\log(1-x)}{x(1-x)}\,dx\right)\\\\ &=\lim_{\epsilon\to 0}\left(\frac{-\log^2(\epsilon)}{1-\epsilon}-2\int_0^{1-\epsilon}\frac{\log(1-x)}{1-x} -2\int_0^1\frac{\log(1-x)}{x}\,dx\right)\\\\ &=\lim_{\epsilon\to 0}\left(-\frac{\log^2(\epsilon)}{1-\epsilon}+2\int_0^{1-\epsilon}\frac12\frac{d\log^2(1-x)}{dx}\,dx -2\int_0^1\frac{\log(1-x)}{x}\,dx\right)\\\\ &=-2\int_0^1\frac{\log(1-x)}{x}\,dx\\\\ &=\frac{\pi^2}{3} \end{align}$$

as was to be shown!

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    $\begingroup$ Would the cowardly down voter care to comment? $\endgroup$ – Mark Viola Jan 23 '17 at 5:04
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    $\begingroup$ Hi Mark ! Don't be too optimistic. He will not. Nice solution by the way $+1$ $\endgroup$ – Claude Leibovici Jan 23 '17 at 6:50
  • $\begingroup$ Thanks Claude! Happy New Year, if I haven't said so. $\endgroup$ – Mark Viola Jan 23 '17 at 15:20
  • $\begingroup$ @Dr.MV, very well. I will refrain from editing your posts. $\endgroup$ – Yuriy S Jan 23 '17 at 17:30
  • $\begingroup$ @YuriyS Thank you Yuriy. Much appreciative. -Mark $\endgroup$ – Mark Viola Jan 23 '17 at 17:34

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