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Find the residue at $z=-2$ for

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

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 of

$g(z)$ at $z=-2$

The problem is I cant ever get a factor of $\frac{1}{z+2}$ what should I do?

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  • $\begingroup$ Write $(z+1)^k=((z+2)-1)^k$ and use the binomial theorem... $\endgroup$ – tired Jan 13 '15 at 13:36
  • $\begingroup$ Can you show me how to do that for a general $k$? please? $\endgroup$ – Amad27 Jan 13 '15 at 13:40
  • $\begingroup$ the point is, that you only need it here until $k=3$ because as you already mentioned only the term with $1/(z+2)$ is interesting :) $\endgroup$ – tired Jan 13 '15 at 13:47
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$\psi(z)$ is regular over $\Re(z)>0$, hence $z=-2$ is a triple pole for $g(z)$. This gives: $$\operatorname{Res}\left(g(z),z=-2\right)= \frac{1}{2}\left.\frac{d^2}{dz^2} (z+2)^3 g(z)\right|_{z=-2}=\frac{1}{2}\left.\frac{d^2}{dz^2} \frac{\psi(-z)}{z+1}\right|_{z=-2}\tag{1}$$ so: $$\operatorname{Res}\left(g(z),z=-2\right)=\left.\frac{\psi(-z)}{(1+z)^3}+\frac{ \psi'(-z)}{(1+z)^2}+\frac{\psi''(-z)}{2(1+z)}\right|_{z=-2} = -\psi(2)+\psi'(2)-\frac{\psi''(2)}{2}$$ leading to: $$\begin{eqnarray*}\operatorname{Res}\left(g(z),z=-2\right)&=& -(1-\gamma)+\left(1-\zeta(2)\right)-\frac{\psi''(2)}{2}\\&=&\gamma-\zeta(2)+\sum_{n\geq 0}\frac{1}{(n+2)^3}\\&=&\color{red}{\gamma-\zeta(2)+\zeta(3)-1}.\tag{2}\end{eqnarray*}$$

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  • $\begingroup$ I am trying to avoid the limit approach. Thanks anyway!This derivative method is too cumbersome without a calculator. How did you calculate $\psi"(2)$ without a calculator? $\endgroup$ – Amad27 Jan 13 '15 at 13:57
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    $\begingroup$ I will accept the answer because this is a nice method but otherwise, we could use the taylor series representation for $\psi(z + 1)$ $\endgroup$ – Amad27 Jan 13 '15 at 14:16
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    $\begingroup$ He was integrating $g(z)$ along a circle centered in the right point. Nothing wrong. By the way, why do you have two accounts? $\endgroup$ – Jack D'Aurizio Jan 13 '15 at 15:38
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    $\begingroup$ You are amazing, upvotes * 1 million! but which law or property states this? Ive never seen this definition/ $\endgroup$ – Amad27 Jan 14 '15 at 14:05
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    $\begingroup$ Ok nice! I used the series representation, the coefficient of $\frac{1}{z-n}$ is $1$ The question remains, what is the residue of $$\frac{(\psi(-z) + \gamma)^2}{(z+1)(z+2)^3}$$ at $z=-2$ I will try it, and post a question later. $\endgroup$ – Amad27 Jan 14 '15 at 14:32

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