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

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SUMMARY

The residue of the function g(z) = ψ(-z)/((z+1)(z+2)^3) at z = -2 is determined using the Laurent expansion. The digamma function ψ(-z) can be expressed as ψ(-z) = -γ - Σ(-1)^kζ(k+1)(z+1)^k, where ζ(k) is the Riemann-Zeta function. To find the residue, one must identify the coefficient of 1/(z+2) in the series expansion, which is complicated by the pole of order 3 at z = -2. The residue calculation requires careful consideration of singularities and may involve analytic continuation techniques.

PREREQUISITES
  • Understanding of Laurent series and residue theory in complex analysis.
  • Familiarity with the digamma function ψ(z) and its properties.
  • Knowledge of the Riemann-Zeta function ζ(z) and its series representation.
  • Experience with analytic continuation and singularity analysis in complex functions.
NEXT STEPS
  • Study the properties and singularities of the digamma function ψ(z) in detail.
  • Learn about residue calculation techniques for higher-order poles in complex analysis.
  • Explore analytic continuation methods for extending functions beyond their initial domains.
  • Investigate the application of series expansions in evaluating complex integrals.
USEFUL FOR

Mathematicians, physicists, and students specializing in complex analysis, particularly those focused on residue theory and special functions like the digamma and Riemann-Zeta functions.

Amad27
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Homework Statement


Find the residue at z=-2 for

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

Homework Equations



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

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 can't ever get a factor of $$\frac{1}{z+2}$$ because of the cubed (z+2) in the bottom. What should I do?
 
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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)##?
 
Amad27 said:

Homework Statement


Find the residue at z=-2 for

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

Homework Equations



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

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 can't ever get a factor of $$\frac{1}{z+2}$$ because of the cubed (z+2) in the bottom. What should I do?

First: be very careful when using a series expansion such as
\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k
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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