Definite integral from 0 to 1 of : ln(x)ln(1-x)dx

1. Aug 15, 2006

Gagle The Terrible

I would really like to post the work I did, but it is gibberish !
I don't know how to tackle this integral :
definite integral from 0 to 1 of : ln(x)ln(1-x)dx

The "traditionnal" methods don't work but I assure you that I have tried much more !

2. Aug 16, 2006

benorin

That integral is, on cursorary examination, sorta nasty

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced [Broken] told me that

$$\int_{0}^{1}\log x\log (1-x) \, dx=2-\frac{\pi ^2}{6}$$

so I expect some quick manipulation to produce the Basel problem (proving that $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi ^2}{6}$,) a computer also told me that

$$\int\log x\log (1-x) \, dx = (x\log x-x)\log (1-x)-x\log x+2x+\mbox{Li}_2 (x)+\log (-1+x)$$

where $$\mbox{Li}_2 (x)=\sum_{k=1}^{\infty}\frac{x^k}{k^2}$$ is the dilogarithm. The dilog gets us to the Basel problem since $$\mbox{Li}_2 (1)=\frac{\pi^2}{6}$$ and hey, there's even a handy formula for the dilog, namely

$$\mbox{Li}_2 (t)=-\int_{0}^{t}\frac{\log (1-x)}{x}dx$$

and I'd conjecture that you can pull this off with integration by parts...

-Ben

Last edited by a moderator: May 2, 2017
3. Aug 16, 2006

WigneRacah

We have
$$I = \int_0^1 \ln x \ln (1-x) dx = \lim_{a\rightarrow 1} \int_0^a \ln x \ln (1-x) dx$$.

Integrating by part

$$I= \lim_{a \rightarrow 1}( - \ln (1-a) + \int_0^a \frac{x \ln x}{1-x} - \int_0^a \frac{x}{1-x})$$.

By performing such a limit we have

$$I= 2 + \int_0^1 \frac{\ln x}{1-x} dx$$.

Now from the substitution $x = e^{-u}$ and by remembering the integral expression for the $\zeta$ function:

$$\zeta(s)= \frac{1}{\Gamma (s)} \int_0^\infty \frac{x^{s-1}}{e^x -1 } d x$$

we get

$$I= 2 - \zeta (2) = 2 - \frac{\pi^2}{6}$$.

Last edited: Aug 16, 2006
4. Aug 16, 2006

benorin

in $$\int_{0}^{1}\log x\log (1-x) \, dx$$ do integration by parts with $$u=\log x$$ so that $$du=\frac{dx}{x}$$ and $$dv=\log (1-x) dx$$ so that $$v=(1-x)\left( 1-\log (1-x)\right)$$ to get

$$\int_{0}^{1}\log x\log (1-x) \, dx = \left[ (1-x)\log x\left( 1-\log (1-x)\right) \right]_{x=0}^{1}- \int_{0}^{1}(1-x)\left( 1-\log (1-x)\right)\frac{dx}{x}$$
$$=\left[ (1-x)\log x\left( 1-\log (1-x)\right) \right]_{x=0}^{1}- \int_{0}^{1}\left[ \frac{1}{x}-1-\frac{\log (1-x)}{x}+\log (1-x) \right] \, dx$$
$$=\left[ (1-x)\log x\left( 1-\log (1-x)\right) -\log x +x -(1-x)\left( 1-\log (1-x)\right) \right]_{x=0}^{1}-\mbox{Li}_2 (1)$$

now it is a matter of taking limits at x=0 and at x=1 and putting in the value $$\mbox{Li}_2(1)=\frac{\pi^2}{6}$$

5. Aug 17, 2006

dextercioby

$$I=\lim_{b\rightarrow 1,a\rightarrow 0}\int_{a}^{b}\ln x\ln \left( 1-x\right) dx=...?$$

One knows that

$$\ln \left( 1-x\right) =-\sum_{k=1}^{\infty }\frac{x^{k}}{k}.$$

for x smaller than 1. So

$$I=\lim_{b\rightarrow 1,a\rightarrow 0}\int_{b}^{a}\ln x\left( \sum_{k=1}^{\infty }\frac{x^{k}}{k}\right) dx=\sum_{k=1}^{\infty }\frac{1}{k}\left(\lim_{b\rightarrow 1,a\rightarrow 0}\int_{b}^{a}x^{k}\ln xdx\right) =...=\sum_{k=1}^{\infty }\frac{1}{k\left( k+1\right) ^{2}}.$$

To evaluate the last sum, use that

$$1=k+1-k$$

and the definition of the Riemann zeta function. U'll easily get $2-\zeta(2)$ you were supposed to get.

Daniel.

P.S. Don't worry about pulling sum out of the integral sign and passing sum symbol over the limit sign. It usually works.

Last edited: Aug 17, 2006
6. Aug 17, 2006

dextercioby

To get a better grip of what i did, try to show that

$$\int_{0}^{\infty} \frac{x}{e^{x}+1} \ dx =\eta(2) =\frac{\pi^{2}}{12}$$

, where \eta is Dirichlet's eta function.

Use the integral above to compute the integrals on page 154 from Ch.Kittel's "Introduction to Solid State Physics", 7-th edition.

Daniel.

Last edited: Aug 17, 2006