I am trying to compute $\int_c(z^2)(z-1)/\sin^2(\pi z) $ where $c$ is the closed circle centered at origin with radius $3/2.$
Am trying to compute the Laurent series for $\sin^2(\pi z)$ to establish the order of the poles of integrand, however am stuck with a Taylor expansion in the denominator. (that is the term with power $2$ )
Is there a trick for flipping it up to the numerator and getting rid of the term with power $2$?
Thanks
The double pole in $\sin^{-2}(\pi z)$ at $z=0$ is cancelled by the factor $z^2$, so the residue at $z=0$ is $0$.
The double pole at $z=1$ is turned into a simple pole by the factor $z-1$, so no series calculation is required here; since $\sin\pi x= \pi x+O\left(x^3\right)$, the residue is just $\frac1{\pi^2}$ times the value of $z^2$ at $z=1$, which is $1$.
The double pole at $z=-1$ has no matching factor in the numerator, so the integrand has a double pole at $z=-1$. But note that the Maclaurin series for $\sin\pi z$ at $z=-1$ only contains odd terms, so its square only contains even terms, beginning with $(z+1)^2$. If we take the reciprocal of this series, it still only has even terms, beginning with exponent $-2$. Thus the only part of the series that we need for the residue is the coefficient of the $(z+1)^{-2}$ term, which is again $\frac1{\pi^2}$ without the need for any series calculation. This needs to be multiplied by the coefficient of $(z+1)^1$ in the numerator, that is, the derivative of the numerator at $z=-1$, which is $5$.
Thus, altogether the residue is $\frac6{\pi^2}=\frac1{\zeta(2)}$.