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Hard problem

  1. Oct 8, 2006 #1
    Here is my problem:

    I need to integrate:

    [tex] (\frac{sin \alpha z}{\alpha z})^2\frac{\pi}{sin\pi z}[/tex]

    around a circle of large radii and prove:

    [tex]\sum_{m=1}^\infty(-1)^{n-1}


    (\frac{sin m\alpha}{m\alpha})^2


    =\frac{1}{2}[/tex]

    I'm kind stumped.

    I've been looking at books for a while now and the only useful things i've discovered are:

    [tex]\frac{\pi}{sin \pi z}=\Gamma(z)\Gamma(1-z)[/tex]

    and also

    [tex] \frac{sin z}{z}=\sum_{n=0}^\infty C_n z^{2n-1}[/tex]

    Can anyone help me out?

    Thanks!
     
    Last edited: Oct 8, 2006
  2. jcsd
  3. Oct 9, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well,
    [tex] (\frac{sin \alpha z}{\alpha z})^2\frac{\pi}{sin\pi z}[/tex]
    has poles at every integer, since sine will be 0 there.

    The integral will be an infinite sum of residues.
     
    Last edited: Oct 9, 2006
  4. Oct 9, 2006 #3
    Thanks Halls, I just got all pleased with myself then realized I've hit another snag:

    Ok, so I just used the rule:

    [tex]Res=\frac{g(z_0))}{h(z_0)'}[/tex]

    With :

    [tex]g(z)=(\frac{sin(\alpha z)}{\alpha z})^2[/tex]

    and

    [tex]g(z)=\frac{1}{\pi}sin(\pi z)[/tex]

    and so

    [tex]g'(z)=cos(\pi z)[/tex]

    Clearly there are zeros for m=0,1,2...and so the when [tex]z\rightarrow z_0[/tex]

    we end up with the following residues:

    [tex]\sum_{m=1}^\infty(-1)^{n-1} (\frac{sin m\alpha}{m\alpha})^2[/tex]

    Now, the integral is [tex]2\pi i \sum Res[/tex]

    Can you help me figure out how to get:

    [tex]\sum_{m=1}^\infty(-1)^{n-1} (\frac{sin m\alpha}{m\alpha})^2=\frac{1}{2}[/tex]

    Essentially I need to figure how to get rid of the 'i' also figure out what the integral goes to...

    Thanks!
     
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