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

  • #1
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
 

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  • #2
benorin
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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

[tex]\int_{0}^{1}\log x\log (1-x) \, dx=2-\frac{\pi ^2}{6}[/tex]

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

[tex]\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)[/tex]

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

[tex]\mbox{Li}_2 (t)=-\int_{0}^{t}\frac{\log (1-x)}{x}dx[/tex]

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

-Ben
 
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  • #3
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We have
[tex]I = \int_0^1 \ln x \ln (1-x) dx = \lim_{a\rightarrow 1} \int_0^a \ln x \ln (1-x) dx[/tex].

Integrating by part

[tex]I= \lim_{a \rightarrow 1}( - \ln (1-a) + \int_0^a \frac{x \ln x}{1-x} - \int_0^a \frac{x}{1-x})[/tex].

By performing such a limit we have

[tex]I= 2 + \int_0^1 \frac{\ln x}{1-x} dx [/tex].

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

[tex]\zeta(s)= \frac{1}{\Gamma (s)} \int_0^\infty \frac{x^{s-1}}{e^x -1 } d x[/tex]

we get

[tex]I= 2 - \zeta (2) = 2 - \frac{\pi^2}{6}[/tex].
 
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  • #4
benorin
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in [tex]\int_{0}^{1}\log x\log (1-x) \, dx[/tex] do integration by parts with [tex]u=\log x[/tex] so that [tex]du=\frac{dx}{x}[/tex] and [tex]dv=\log (1-x) dx[/tex] so that [tex]v=(1-x)\left( 1-\log (1-x)\right) [/tex] to get

[tex]\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}[/tex]
[tex]=\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[/tex]
[tex]=\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)[/tex]

now it is a matter of taking limits at x=0 and at x=1 and putting in the value [tex]\mbox{Li}_2(1)=\frac{\pi^2}{6}[/tex]
 
  • #5
dextercioby
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[tex] I=\lim_{b\rightarrow 1,a\rightarrow 0}\int_{a}^{b}\ln x\ln \left( 1-x\right) dx=...?[/tex]

One knows that

[tex] \ln \left( 1-x\right) =-\sum_{k=1}^{\infty }\frac{x^{k}}{k}.[/tex]

for x smaller than 1. So

[tex]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}}. [/tex]

To evaluate the last sum, use that

[tex] 1=k+1-k [/tex]

and the definition of the Riemann zeta function. U'll easily get [itex] 2-\zeta(2) [/itex] 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.
 
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  • #6
dextercioby
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To get a better grip of what i did, try to show that

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

, 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.
 
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