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Math Help - Lebesgue integration questions

  1. #1
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    Lebesgue integration questions

    I have a couple of integrals that I need to evaluate - they are in the context of Lebesgue integrals. They seem similar to questions regarding contour integration that I've done before, but I'm not sure if that applies here.

    I need to show the following:

    1) \int_{0}^{\infty}\frac{\sin(\alpha x)}{e^x-1}dx=\sum_{n=1}^{\infty}\frac{\alpha}{\alpha^2 + n^2}

    2) \int_{0}^{1}\frac{\log x}{1+x}dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}

    Thanks for any help!
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  2. #2
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    Quote Originally Posted by angela87 View Post
    I have a couple of integrals that I need to evaluate - they are in the context of Lebesgue integrals. They seem similar to questions regarding contour integration that I've done before, but I'm not sure if that applies here.

    I need to show the following:

    1) \int_{0}^{\infty}\frac{\sin(\alpha x)}{e^x-1}dx=\sum_{n=1}^{\infty}\frac{\alpha}{\alpha^2 + n^2}

    2) \int_{0}^{1}\frac{\log x}{1+x}dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}

    Thanks for any help!
    For the first one multiply the numerator and denomiantor of the integrand by e^{-x}

    This gives

    \int_{0}^{\infty}\frac{e^{-x}\sin(\alpha x)}{1-e^{-x}}dx

    Note that e^{-x} < 1 for all x \in (0,\infty)

    Expand the denomiantor in a geometric series to get

    \int_{0}^{\infty}e^{-x}\sin(\alpha x) \sum_{n=0}^{\infty}e^{-nx}dx=\sum_{n=0}^{\infty}\sin(\alpha x)e^{-(n+1)x}dx

    Integrate by parts to get

    \sum_{n=0}^{\infty}\frac{\alpha}{(n+1)^2+\alpha}

    Just reindex the sereis and your done Note that some of the above steps still need formal justification.

    For the 2nd one it is a similar trick

    write \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n

    Then use integration by parts again.

    Good luck
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  3. #3
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    good show...i had just started looking into bessel functions to get the first one. i guess the wisdom of that would depend on knowing whether this problem set came from an undergrad course or not. anyway, nice moves.
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  4. #4
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    Quote Originally Posted by angela87 View Post
    2) \int_{0}^{1}\frac{\log x}{1+x}dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}
    we have \int_{0}^{1}{x^{n}\ln x\,dx}=\int_{0}^{1}{\int_{1}^{x}{\frac{x^{n}}{t}\,  dt}\,dx}=-\int_{0}^{1}{\int_{0}^{t}{\frac{x^{n}}{t}\,dx}\,dt  }=-\frac{1}{(n+1)^{2}}, so \int_{0}^{1}{\frac{\ln x}{1+x}\,dx}=\sum\limits_{n=0}^{\infty }{(-1)^{n}\int_{0}^{1}{x^{n}\ln x\,dx}}=\sum\limits_{n=0}^{\infty }{\frac{(-1)^{n+1}}{(n+1)^{2}}}=\sum\limits_{n=1}^{\infty }{\frac{(-1)^{n}}{n^{2}}}, as required.

    but you're gonna have to justify why we can swap sum and integral, so there's your job.
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