Residue of f(z) involving digamma function

In summary, we have discussed finding the residues of a function with simple poles at every positive integer n, and how to express the digamma function as a simple pole for each n. We have also explored finding residues for higher order poles and found that the coefficient can be calculated in a similar manner.

Homework Statement

Find the residue of:

$$f(z) = \frac{(\psi(-z) + \gamma)}{(z+1)(z+2)^3} \space \text{at} \space z=n$$

Where $n$ is every positive integer because those $n$ are the poles of $f(z)$

The Attempt at a Solution

This is a simple pole, however:

$$\lim_{z \to n} \frac{(z-n)\psi(-z)}{(z+1)(z+2)^3}$$

But this gives 0, which is incorrect?

For n to be a simple pole of ##\psi (-z) ## you will need to be able to express ##\psi (-z) = (z+n)g(z)## where g(z) is analytic at z=-n. Be careful with your signs.
Using the expansion ##\psi (n)+\gamma= \sum_{k=1}^n-1 \frac 1k ## shows that for every positive integer n there will be a simple pole at z=-n, since the sum will include k=0.

RUber said:
For n to be a simple pole of ψ(−z)\psi (-z) you will need to be able to express ψ(−z)=(z+n)g(z)\psi (-z) = (z+n)g(z) where g(z) is analytic at z=-n. Be careful with your signs.
Using the expansion ψ(n)+γ=∑nk=1−11k\psi (n)+\gamma= \sum_{k=1}^n-1 \frac 1k shows that for every positive integer n there will be a simple pole at z=-n, since the sum will include k=0.

Yes I see this, but what about higher order poles? consider:

Suppose we are given:

Find the residue at z=n of $$(\psi(-z) + \gamma)^2$$

Where $$\psi(-z)$$ means the digamma function.

$$(\psi(-z) +\gamma)^2 = \left( 1 + \frac{1}{2} + \frac{1}{z} + \frac{1}{3} + \frac{1}{z-1} + \frac{1}{4} + \frac{1}{z-2} + ... + \frac{1}{n} + \frac{1}{z-n} + ... \right)\left( 1 + \frac{1}{2} + \frac{1}{z} + \frac{1}{3} + \frac{1}{z-1} + \frac{1}{4} + \frac{1}{z-2} + ... + \frac{1}{n} + \frac{1}{z-n} + ... \right)$$

We see that we have:

$$\frac{1}{z-n} \cdot \left( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + \frac{1}{z-1} + \frac{1}{z-2} + ... + \frac{1}{z-n} \right)$$

The pure coefficient of $$\frac{1}{z-n}$$ is simply:

$$\left( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right) = H_n$$

Where $$H_n$$ means the Harmonic Number.

But then we also had the other $$\frac{1}{z-n}$$ in the other parentheses so twice this.

$$2 \left( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right) = 2H_n$$

But does finding the coefficient work for higher order residues?

Thanks!

What is the definition of the residue of f(z) involving digamma function?

The residue of f(z) involving digamma function is the coefficient of the term with (z-a)^-1 in the Laurent expansion of the function f(z) at the point a.

How is the residue of f(z) involving digamma function calculated?

The residue of f(z) involving digamma function can be calculated by using the formula Res(f; a) = lim(z->a) (z-a)f(z), where a is the point of interest.

What is the significance of the residue of f(z) involving digamma function in complex analysis?

The residue of f(z) involving digamma function is used to calculate the value of certain complex integrals, which have applications in various areas of mathematics, physics, and engineering.

Are there any special cases or exceptions for calculating the residue of f(z) involving digamma function?

Yes, there are some special cases where the residue of f(z) involving digamma function may not exist or may need to be handled differently. These cases include poles of higher order and essential singularities.

How is the residue of f(z) involving digamma function related to the Cauchy integral formula?

The residue of f(z) involving digamma function is closely related to the Cauchy integral formula, which states that the value of a complex integral is equal to 2πi times the sum of residues of all singularities enclosed by the contour of integration.

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