- #1

- 39

- 0

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

- Thread starter Gagle The Terrible
- Start date

- #1

- 39

- 0

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

- #2

- 1,278

- 82

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

- 34

- 0

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].

[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

- 1,278

- 82

[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

- 13,002

- 551

[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.

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

- 13,002

- 551

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.

[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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