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I have to calculate the sum of the residues at the $n$ m-order poles of a complex function $$f(z)=\frac{z^{np-1}}{(az^n+b)^m}\ln z$$ when applying Residue Theorem for an integral. In this function, $m,n,p$ are positive integers, and $m>p$. $a,b\in\mathbf{R}$ are positive.

When using the limit equation for the residue of a m-order pole, I find it hard to get the high order derivative of this function.

Can anyone help me? Thanks for any guidance.

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  • $\begingroup$ You could rewrite the expression as a Laurent Series in $a/b$ $\endgroup$ – C. Brendel Nov 13 '19 at 5:12
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You want the sum of the residues of the poles at radius $b/a$. These lie on an annulus (with inner radius $b/(2a)$ and outer radius $2b/a$, say). The positively oriented path integral along its inner boundary component encloses only the (potential) singularity at $z = 0$; call this $A$. The positively oriented path integral along its outer boundary component encloses all $n$ of the $m^\text{th}$-order poles you are interested in and the singularity at $z = 0$; call this $B$. The quantity you want is $\frac{1}{2\pi\mathrm{i}}(B - A)$. However, $B$ is hard to calculate. The negatively oriented path integral along the annulus's outer boundary component encloses only the (potential) singularity at $z = \infty$; call this integral $C$. Note that $C = -B$, so the sum of residues you want is $\frac{1}{2\pi\mathrm{i}}(-C-A)$.

I observe the direct attack, computing residues of all $n$ of the $m^\text{th}$-order pole is hard. Computing the residues of the singularities at $0$ and $\infty$ might be easier...

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  • $\begingroup$ $ln z$ seems to have no residue at z=0… $\endgroup$ – worstcoder Nov 13 '19 at 6:28

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