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Math Help - improper integral

  1. #1
    Super Member Random Variable's Avatar
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    improper integral

    Show that  \displaystyle \int^{\infty}_{0} \frac{\sin x}{x} \ \ln x \ dx = -\gamma \int_{0}^{\infty} \frac{\sin x}{x} \ dx = - \frac{\pi}{2} \ \gamma (where  \gamma is Euler's constant).
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  2. #2
    Super Member Random Variable's Avatar
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    Hint:

    Spoiler:
    Start with the fact that  \displaystyle \int^{\infty}_{0} \frac{ \ln x} {1+x^{2}} \ dx = 0


    Bigger Hint:

    Spoiler:
     \displaystyle \int^{\infty}_{0} e^{-x} \ln x = - \gamma
    Last edited by Random Variable; January 24th 2011 at 10:41 AM.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Show that  \displaystyle \int^{\infty}_{0} \frac{\sin x}{x} \ \ln x \ dx = -\gamma \int_{0}^{\infty} \frac{\sin x}{x} \ dx = - \frac{\pi}{2} \ \gamma (where  \gamma is Euler's constant).
    I would proceed as follows:


    Spoiler:



    Let \displaystyle I(\alpha)=\int_0^{\infty}\sin(x)x^{\alpha-1}\text{ }dx. By previous problem we have that \displaystyle I(\alpha)=\frac{\pi}{2\Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)} for \alpha\in(-1,1) and thus on that interval

    \displaystyle \begin{aligned}I'(\alpha) &=\frac{-\pi\left[-2\Gamma'(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)-\pi\Gamma(1-\alpha)\sin\left(\frac{\pi\alpha}{2}\right)\right]}{\left(2\Gamma\left(1-\alpha\right)\cos\left(\frac{\pi\alpha}{2}\right)\  right)^2}\\ &=\frac{\pi}{4}\frac{2\Gamma'(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)+\pi\Ga  mma(1-\alpha)\sin\left(\frac{\pi\alpha}{2}\right)}{\left  (\Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)\right)  ^2}\end{aligned}

    That said, using Lebniz's differentiation under the integral technique (checking that it's applicable) we find that


    \displaystyle I'(\alpha)=\int_0^{\infty}\frac{\partial}{\partial \alpha}\sin(x)x^{\alpha-1}\text{ }dx=\int_0^\infty} \sin(x)x^{\alpha-1}\log(x)\text{ }dx


    Thus, our integral is \displaystyle I'(0)=\frac{\pi}{4}\cdot2\Gamma'(1)=\frac{-\pi\gamma}{2}.

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  4. #4
    Super Member Random Variable's Avatar
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    It's amazing how I don't realize that many of the problems I post are interconnected until you point it out to me.


     \displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^2} \ dx = 0

     \displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^{2}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} \ln x \ \mathcal{L} \{ \sin t \} (x) \ dt \ dx = 0

     \displaystyle \int^{\infty}_{0} \sin t \int^{\infty}_{0} e^{-xt} \ln x \ dx \ dt =0

      \displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \Big(\int^{\infty}_{0} e^{-u} \ln u \ du - \ln t \int^{\infty}_{0} e^{-u}\ du\Big) \ dt =0

     \displaystyle \int^{\infty}_{0} \frac{\sin t}{t} (-\gamma - \ln t) dt = 0

    so  \displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \ln t \ dt = -\gamma \int^{\infty}_{0} \ \frac{\sin t}{t} \ dt
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    It's amazing how I don't realize that many of the problems I post are interconnected until you point it out to me.


     \displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^2} \ dx = 0

     \displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^{2}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} \ln x \ \mathcal{L} \{ \sin t \} (x) \ dt \ dx = 0

     \displaystyle \int^{\infty}_{0} \sin t \int^{\infty}_{0} e^{-xt} \ln x \ dx \ dt =0

      \displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \Big(\int^{\infty}_{0} e^{-u} \ln u \ du - \ln t \int^{\infty}_{0} e^{-u}\ du\Big) \ dt =0

     \displaystyle \int^{\infty}_{0} \frac{\sin t}{t} (-\gamma - \ln t) dt = 0

    so  \displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \ln t \ dt = -\gamma \int^{\infty}_{0} \ \frac{\sin t}{t} \ dt

    Yeah, but I'm often lazy and just look for the quickest most mechanical solution, you take time and come up with neat non-obvious solutions. So, in the end, which is better?
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  6. #6
    Super Member Random Variable's Avatar
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    Can we justify changing the order of integration, or is my solution invalid?
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  7. #7
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    Solution By use of MAZ Identity

    Last edited by uniquesailor; February 16th 2011 at 01:19 PM.
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  8. #8
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    Quote Originally Posted by uniquesailor View Post
    ...
    What is 'MAZ identity', if you don't mind me asking?
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