Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
5  
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
2  
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

8 Answers 8

up vote 30 down vote accepted

$$\int_0^1\frac{\log t}{1-t}dt=\int_0^1\log(1-u)\frac{du}{u}=\int_0^\infty \log(1-e^{-v})dv =-\frac{\pi^2}{6}.$$ 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: $$\int_0^1\left(\frac{\log t}{1-t}\right)^2dt=\int_0^1\left(\frac{\log(1-u)}{u}\right)^2du=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{nm}\int_0^1 u^{n+m-2}du$$ $$=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{nm(n+m-1)}=\sum_{m=1}^\infty \frac{1}{m^2}+\sum_{n=2}^\infty\frac{1}{n}\frac{1}{n-1}\sum_{m=1}^\infty\left(\frac{1}{m}-\frac{1}{n+m-1}\right)$$ $$=\frac{\pi^2}{6}+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)\sum_{m=1}^n\frac{1}{m}=\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{3}. $$

share|improve this answer
    
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.

share|improve this answer
    
+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.

share|improve this answer

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 $$\sum_{k=1}^{\infty} \left(\frac{1}{k^2} \int_0^{\infty} u^2 e^{-u}du\right) = \left(\int_0^{\infty} u^2 e^{-u}du\right) \left(\sum_{k=1}^{\infty} \frac{1}{k^2}\right) = \Gamma(3) \zeta(2) = \frac{\pi^2}{3}.$$

share|improve this answer
1  
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
3  
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:

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

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

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

share|improve this answer

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 $\displaystyle\sum_{k=1}^\infty \frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}$.

share|improve this answer

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}

share|improve this answer

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}\bracks{\ln\pars{t} \over 1 - t}^{2}\,\dd t ={\pi^{2} \over 3}:\ {\large ?}}$.

\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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.