I'm having trouble finding residue of the function
$$f(z)=\frac{z^5}{\sin\left(\frac{1}{\large{z^2}}\right)}$$
at infinity. Wolfram kindly informs that it is equal to $-\frac{7}{360}$ (and gives expansion). How do get that? Infinity is the pole of $7$th order. It seems not possible to efficiently expand $f(z)$ or $f\left(\frac{1}{z}\right)$ without computers to look for coefficient. Or is it? Is it possible there is some tricky curve such that I'll be able to integrate over it? Thank you.
Update:
Silly me, it's so easy:
$$z^5=\left(a_7 z^7 + \ldots + a_0 + \frac{a_{-1}}{z} + \ldots\right)\left(\frac{1}{z^2}-\frac{1}{6z^6}+\frac{1}{120z^{10}}-\ldots\right)$$
By comparing I get: $$\begin{align} &a_{-1}-\frac{a_3}{6}+\frac{a_7}{120}=0\\ &a_3-\frac{a_7}{6}=0\\ &a_7=1\\ \end{align}$$
And I can easily find $a_{-1}$.
Thank you again, Enigma.