# A proof of $\int_{0}^{1}\left( \frac{\ln t}{1-t}\right)^2\,\mathrm{d}t=\frac{\pi^2}{3}$

What is the proof of the following: $$\int_{0}^{1} \left(\frac{\ln t}{1-t}\right)^2 \,\mathrm{d}t=\frac{\pi^2}{3} \>?$$

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You have got the sign wrong for sure. The integrand is always negative in $(0,1)$. –  user17762 Oct 26 '11 at 0:57
by typing the wrong question. I'll post the correct –  Gardel Oct 26 '11 at 1:02
Wolfram Alpha can do this but I can't figure out how to make it show the steps. –  opt Oct 26 '11 at 1:43

$$\int_0^1\frac{\log t}{1-t}dt=\int_0^1\log(1-u)\frac{du}{u}=\int_0^\infty \log(1-e^{-v})dv =-\frac{\pi^2}{6}.$$ For the last part see an answer of mine here.

For the revised question, substitute $u=1-t$ and expand into a product of Taylor series, then use some of partial fraction decomposition, sum splitting, reindexing, and telescoping properties: $$\int_0^1\left(\frac{\log t}{1-t}\right)^2dt=\int_0^1\left(\frac{\log(1-u)}{u}\right)^2du=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{nm}\int_0^1 u^{n+m-2}du$$ $$=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{nm(n+m-1)}=\sum_{m=1}^\infty \frac{1}{m^2}+\sum_{n=2}^\infty\frac{1}{n}\frac{1}{n-1}\sum_{m=1}^\infty\left(\frac{1}{m}-\frac{1}{n+m-1}\right)$$ $$=\frac{\pi^2}{6}+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)\sum_{m=1}^n\frac{1}{m}=\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{3}.$$

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I'm sorry for posting wrong. Has been updated to correct post –  Gardel Oct 26 '11 at 1:08
@Gardel: I've updated my answer with a solution to the revised question. If you think something's too unclear just tell me. –  anon Oct 26 '11 at 1:40
I like the way you solved the revised question. Good job. –  smanoos Oct 26 '11 at 1:44
@anon, really good solution. Thanks! –  Tapu Oct 26 '11 at 18:33

This was intended to be a comment on Mike Spivey's answer, but alas, it was too long.

For $n>0$,

\begin{align} \int_0^1\left(\frac{\log(t)}{1-t}\right)^n\mathrm{d}t &=\int_0^\infty\left(\frac{-s}{1-e^{-s}}\right)^ne^{-s}\mathrm{d}s\\ &=(-1)^n\int_0^\infty s^ne^{-s}\sum_{k=0}^\infty\binom{-n}{k}(-1)^ke^{-ks}\;\mathrm{d}s\\ &=(-1)^n\Gamma(n+1)\sum_{k=0}^\infty\binom{-n}{k}(-1)^k(k+1)^{-n-1}\\ &=(-1)^n\Gamma(n+1)\sum_{k=0}^\infty\binom{k+n-1}{n-1}(k+1)^{-n-1}\\ &=(-1)^n\Gamma(n+1)\sum_{k=1}^\infty\binom{k+n-2}{n-1}k^{-n-1}\\ &=(-1)^n\Gamma(n+1)\sum_{k=1}^\infty\frac{1}{(n-1)!}\sum_{j=0}^{n-1}\genfrac{[}{]}{0}{0}{n-1}{j}k^jk^{-n-1}\\ &=(-1)^nn\sum_{j=0}^{n-1}\genfrac{[}{]}{0}{0}{n-1}{j}\zeta(n-j+1) \end{align} where $\genfrac{[}{]}{0}{0}{n}{k}$ is a Stirling number of the first kind.

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+1 nice generalization. –  anon Oct 26 '11 at 9:53
+1.. Yes, nice! –  Mike Spivey Oct 26 '11 at 15:35

This is by no means a complete solution but a possible route.

Letting $t = \frac1x$ note that $$I = \int_{0}^{1} \left(\frac{\ln t}{1-t}\right)^2 \,\mathrm{d}t= \int_1^{\infty} \left(\frac{\ln t}{1-t}\right)^2 \,\mathrm{d}t = \int_0^{\infty} \left(\frac{\ln (1+t)}{t}\right)^2 \,\mathrm{d}t$$

Setting $1+t = e^x$, we get $$I = \int_0^{\infty} \frac{x^2}{(e^x-1)^2} e^x dx = \int_0^{\infty} \frac{x^2}{(e^{x/2}-e^{-x/2})^2} dx = 2 \int_{-\infty}^{\infty} \frac{x^2}{\sinh^2(x)} dx$$

The last integral can be done by the method of residues to get $$\int_{-\infty}^{\infty} \frac{x^2}{\sinh^2(x)} dx = \frac{\pi^2}{6}$$ I will fill this in once I get back home.

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Sivaram has shown that $$\int_0^1 \left(\frac{\log t}{1-t}\right)^2 = \int_0^{\infty} \frac{x^2 e^x}{(e^x-1)^2} dx.$$ Here's a different way to complete his argument, plus a generalization in the comments.

If $p = 1 - e^{-x}$, and $Y$ is geometric$(p)$, then $$E[Y] = \frac{1}{p} = \frac{1}{1-e^{-x}} = \frac{e^x}{e^x-1}.$$ But, by definition, $$E[Y] = \sum_{k=0}^{\infty} k(1-p)^{k-1} p = \sum_{k=1}^{\infty} k(e^{-x})^{k-1} (1-e^{-x}) = \sum_{k=1}^{\infty} k(e^{-x})^k (e^x-1).$$ Thus we have $$\int_0^{\infty} \frac{x^2 e^x}{(e^x-1)^2} dx = \int_0^{\infty} x^2 \left(\sum_{k=1}^{\infty} k(e^{-x})^k \right)dx = \sum_{k=1}^{\infty} \left(k \int_0^{\infty} x^2 e^{-kx}dx\right).$$ Finally, if we let $u = kx$, we get $$\sum_{k=1}^{\infty} \left(\frac{1}{k^2} \int_0^{\infty} u^2 e^{-u}du\right) = \left(\int_0^{\infty} u^2 e^{-u}du\right) \left(\sum_{k=1}^{\infty} \frac{1}{k^2}\right) = \Gamma(3) \zeta(2) = \frac{\pi^2}{3}.$$

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Replacing $2$ with $s$ in the last two lines of the argument yields the generalization $$\Gamma(s+1)\zeta(s) = \int_0^{\infty} \frac{x^s \, e^x}{(e^x-1)^2} dx.$$ –  Mike Spivey Oct 26 '11 at 6:18
This is the analog of @anon's method for $\int_0^1\frac{\log t}{1-t}dt$ –  robjohn Oct 26 '11 at 7:59
I completed your generalization, but it was too long to fit in the margin. :-) –  robjohn Oct 26 '11 at 9:51
@robjohn Isn't it also true that $$\int_0^\infty \frac{x^s}{e^x-1}dx = \Gamma(s) \zeta(s)$$? –  Pedro Tamaroff Mar 4 '12 at 1:52
...the $s$ should be $s-1$ –  Pedro Tamaroff Mar 4 '12 at 2:00

A short proof:

Integration by parts:

$$\int_0^1 \left(\frac{\ln(t)}{1-t}\right)^2 \mathrm dt=\left[\left(\frac{1}{1-t}-1\right)\ln(t)^2 \right]_0^1-2\int_0^1\frac{\ln(t)}{1-t} \mathrm dt=-2\int_0^1\frac{\ln(t)}{1-t} \mathrm dt$$

$$\int_0^1\frac{\ln(t)}{1-t} \mathrm dt =-\frac{\pi^2}{6}$$

$$\int_0^1 \left(\frac{\ln(t)}{1-t}\right)^2 \mathrm dt=\frac{\pi^2}{3}$$

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Consider $$\sum_{k=0}^\infty t^k=\frac1{1-t}.\tag1$$ Differentiating $(1)$ with respect to $t$ yields $$\sum_{k=1}^\infty kt^{k-1}=\frac1{(1-t)^2}.\tag2$$ Multiplying both sides of $(2)$ by $\ln^2t$ and integrating from $t=0$ to $t=1$ yields \begin{align} \int_0^1\left(\frac{\ln t}{1-t}\right)^2\ dt&=\int_0^1\sum_{k=1}^\infty kt^{k-1}\ln^2t\ dt\\ &=\sum_{k=1}^\infty k\int_0^1 t^{k-1}\ln^2t\ dt.\tag3 \end{align} Using formula $$\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots\tag4$$ then $(3)$ becomes \begin{align} \int_0^1\left(\frac{\ln t}{1-t}\right)^2\ dt &=2\sum_{k=1}^\infty \frac{1}{k^2}\\ &=\large\color{blue}{\frac{\pi^2}{3}},\tag{Q.E.D.} \end{align} where $\displaystyle\sum_{k=1}^\infty \frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}$.

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I apologise for this extremely late solution, but this is another possible way to evaluate the integral. Consider \begin{align} I(a) &=\int^1_0\frac{x^a}{(1-x)^2}dx\\ &=\sum_{n \ge 0}(n+1)\int^1_0x^{a+n}dx\\ &=\sum_{n \ge 1}\frac{n}{a+n} \end{align} Hence \begin{align} I''(a)=\sum_{n \ge 1}\frac{2n}{(a+n)^3}\\ \end{align} and \begin{align} \int^1_0\frac{\ln^2{x}}{(1-x)^2}dx &=I''(0)\\ &=2\sum_{n \ge 1}\frac{1}{n^2}\\ &=\frac{\pi^2}{3} \end{align}

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Your $I(a)$ diverges. –  xpaul Dec 9 '14 at 22:54


\begin{align}&\color{#66f}{\large\int_{0}^{1}% \bracks{\ln\pars{t} \over 1 - t}^{2}\,\dd t} =\lim_{\epsilon\ \to\ 0^{+}}\int_{\epsilon}^{1}% {\ln^{2}\pars{t} \over \pars{1 - t}^{2}}\,\dd t \\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{% -\,{\ln^{2}\pars{\epsilon} \over 1 - \epsilon} -\int_{\epsilon}^{1}{1 \over 1 - t}\,2\ln\pars{t}\,{1 \over t}\,\dd t} \\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{% -\,{\ln^{2}\pars{\epsilon} \over 1 - \epsilon} -2\int_{\epsilon}^{1}{\ln\pars{t} \over 1 - t}\,\dd t -2\int_{\epsilon}^{1}{\ln\pars{t} \over t}\,\dd t} \\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{% -\,{\ln^{2}\pars{\epsilon} \over 1 - \epsilon} -2\int_{\epsilon}^{1}{\ln\pars{t} \over 1 - t}\,\dd t +2\ln^{2}{\epsilon}} =-2\int_{0}^{1}{\ln\pars{1 - t} \over t}\,\dd t \\[3mm]&=2\int_{0}^{1}{\rm Li}_{2}'\pars{t}\,\dd t =2\,{\rm Li}_{2}\pars{1}=2\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}} _{\ds{=\ \color{#c00000}{\pi^{2} \over 6}}}\ =\ \color{#66f}{\Large{\pi^{2} \over 3}} \approx {\tt 3.2899} \end{align}

$\ds{{\rm Li_{s}}\pars{z}}$ is the PolyLogarithm Function and we used well known properties of them as reported in the above cited link.

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