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

[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 !

Please help

 

[benorin]

That integral is, on cursorary examination, sort of nasty

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced 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

 

[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}.

 

[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}

 

[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.

 

[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.