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Homework Help: Complex analysis - partial fraction expansion

  1. Oct 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that:
    Ʃ(-1)n/(n^2+a^2) (from n=0 to ∞) = pi/[asinh(pi*a)], a[itex]\neq in[/itex], n[itex]\in Z[/itex].

    2. Relevant equations

    f(z) = f(0) + Ʃbn(1/(z-an)+1/an) (from n=1 to ∞) , where bn is the residue of f(z) at an.

    3. The attempt at a solution

    The main problem is I don't how to pick the function f(z) to start with. I am thinking is pi*cot(pi*z) a good starting point? But in this case, we are dealing with the hyperbolic function, so coth will be a better choice??

    Any help is appreciated!
  2. jcsd
  3. Oct 18, 2013 #2


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    First of all the sum is over [itex]n \in \mathbb{Z}[/itex], and the correct statement is
    [tex]\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+a^2}=\frac{\pi}{a \sinh(a \pi)}.[/tex]
    Now find a function with residua [itex](-1)^n[/itex] at [itex]z=n[/itex] for [itex]n \in \mathbb{Z}[/itex].

    You are close with the choice of your function, but the one you gave has residua [itex]+1[/itex]. Multiply it with an entire function which takes the values [itex](-1)^n[/itex] at [itex]z=n \in \mathbb{Z}[/itex]. Then choose an appropriate contour to integrate over and finally deform it appropriately to use the residue theorem.
  4. Oct 18, 2013 #3
    From my notes, i found that Ʃ(-1)nf(n) (from -∞ to ∞) = - Ʃ Res(pi*csc(pi*z), zj) (poles zj of f(z)).

    So, should I let f(n) = 1/(n2+a2) in this case?
  5. Oct 18, 2013 #4
    I've been trying myself to sort out this Laurent series stuff, but I'm pretty sure I'm not following this. Would you mind answering some questions?

    Are we trying to find a function g(z) to expand in a Laurent series which will match the summation at z = n?

    I understand that any appropriate g would have simple poles at ±ai, so would have just the ##a_{-1}## term in its Laurent series. But why would this residue necessarily show up in the other terms of the Laurent series?

    Does this Laurent series converge to g everywhere? Somehow I'm thinking that the series around ai is good only down to -ai -- that is converges in a radius of 2a around the point +ai.

    Or am I completely on the wrong track here?

    If you can explain, I would very much appreciate it.
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