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.
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...