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Find the residue of g(z) at z=-2 using Laurent Expansion

  1. Jan 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the residue at z=-2 for

    $$g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$



    2. Relevant equations

    $$\psi(-z)$$ represents the digamma function, $$\zeta(z)$$ represents the Riemann-Zeta-Function.

    3. The attempt at a solution
    I know that:

    $$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$

    Let $$z \to -1 - z$$ to get:

    $$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$

    Therefore we divide by the other part to get:

    $$\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}$$

    I have to somehow get the coefficient of $$\frac{1}{z+2}$$ because I want to evaluate the residue at z=-2

    The problem is I cant ever get a factor of $$\frac{1}{z+2}$$ because of the cubed (z+2) in the bottom. What should I do?
     
  2. jcsd
  3. Jan 13, 2015 #2

    RUber

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    The residue for a pole of order m at k is given by:
    ##lim _{z->k} [\frac{1}{(m-1)!} \frac{ d^{m-1}}{dz^{m-1}} ((z-k)^m f(z))]##
    For this case, what is m? k = -2. Do you know the derivatives of ##\psi(z)##?
     
  4. Jan 13, 2015 #3

    Ray Vickson

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    First: be very careful when using a series expansion such as
    [tex] \psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k [/tex]
    This may have limited applicability (that is, may be valid only in a subset of ##\mathbb{C}##) and may need to be extended to other regions by some standard techniques of "analytic continuation". Basically, you need to know whether or not ##\psi(-z)## has singularities, and where they are. For example, can it have a pole at ##z = -2##? If so, that would change the residue.

    So, your first order of business should be to understand fully the nature of the singularities of ##g(z)##, at least near ##z = -2##.
     
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