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Math Help - definite integral #4

  1. #1
    Super Member Random Variable's Avatar
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    definite integral #4

     \displaystyle \int^{\pi/2}_{0} x^{2} \cot x \ dx


    I have a solution using contour integration. You could either try to replicate that solution, or preferably find a simpler approach.

    Spoiler:
    Integrate  \displaystyle f(z) = \frac{z^{2}}{e^{z}-1} around the proper contour in the complex plane.
    Last edited by Random Variable; February 14th 2011 at 06:42 PM.
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  2. #2
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    Integrate by parts and the integral equals -2\displaystyle\int_0^{\frac\pi2}x\ln(\sin x)\,dx, now use \displaystyle\ln (\sin x)=-\ln 2-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}, and that gives the answer \dfrac18(2\pi^2\ln2-7\zeta(3)).
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Krizalid View Post
    Integrate by parts and the integral equals -2\displaystyle\int_0^{\frac\pi2}x\ln(\sin x)\,dx, now use \displaystyle\ln (\sin x)=-\ln 2-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}, and that gives the answer \dfrac18(2\pi^2\ln2-7\zeta(3)).
    From where is that series coming?
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  4. #4
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    \begin{aligned}<br />
   \ln (\cos x)&=\text{Re}\ln \left( \frac{1+{{e}^{2ix}}}{2{{e}^{2ix}}} \right) \\ <br />
 & =\text{Re}\left( \ln \left( 1+{{e}^{2ix}} \right)-2ix-\ln 2 \right) \\ <br />
 & =\text{Re}\left( -\sum\limits_{j=1}^{\infty }{\frac{{{(-1)}^{j}}{{e}^{2ijx}}}{j}} \right)-\ln 2 \\ <br />
 & =-\sum\limits_{j=1}^{\infty }{\frac{{{(-1)}^{j}}\cos (2jx)}{j}}-\ln 2, \\ <br />
\end{aligned}

    now put x\mapsto\dfrac\pi2-x and get

    \displaystyle\ln (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\ln 2.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Integrate by parts and the integral equals -2\displaystyle\int_0^{\frac\pi2}x\ln(\sin x)\,dx, now use \displaystyle\ln (\sin x)=-\ln 2-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}, and that gives the answer \dfrac18(2\pi^2\ln2-7\zeta(3)).
    I'm not as quite accomplished as you Kriz, but couldn't we have saved all this trouble and used, for example, a power series solution right from the start? Wouldn't this then give us two inequivalent series with the same sum?

    -Dan
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  6. #6
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    You mean power series for contangent? I don't know it.

    Perhaps it's better you to share your solution.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    You mean power series for contangent? I don't know it.

    Perhaps it's better you to share your solution.
    Okay, okay! (chuckles) I guess I didn't think that one through very far, did I?

    -Dan
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Krizalid View Post
    You mean power series for contangent? I don't know it.

    Perhaps it's better you to share your solution.
    It lurks about halfway down >>this<< page.

    CB
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Krizalid View Post
    You mean power series for contangent? I don't know it.

    Perhaps it's better you to share your solution.
     \displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n}

     \displaystyle \implies \frac1x-\pi\cot(\pi x) = 2\sum_{n=1}^\infty \zeta(2n)x^{2n-1} \quad\quad \text{(for } |x|<1\text{)}
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