# Residue calculation complex analysis

I started to calculate an integral $$\int_{|z| = 106} \cos\left(\frac{1}{6z+1}\left(\frac{1}{(z-1)(z-2)\cdots(z-105)}\right)\right)\,dz$$. I think that a good way is to think about $$\infty$$ as a singular point and investigate its type. If I put $$\infty$$ to the equation, I will get a zero, so its type is zero. And now I need use this technique to calculate the integral.

• Just a thought, but maybe you can use $\cos(1/f(z))=1-\frac{1}{2f(x)^2}+\cdots$. Commented Jun 12, 2020 at 13:29
• It looks interesting. We asked teacher nd he said that first of all we should find out a type of singularity $\infty$.
– lida
Commented Jun 12, 2020 at 14:23

Yes, residue-at-infinity is the simplest approach (not sure how you do it).

The same is done by $$z=1/w$$: the given integral is $$\int_{|w|=1/n}\cos\left(\frac{w}{6+w}\prod_{k=1}^{n-1}\frac{w}{1-kw}\right)\frac{dw}{w^2}\qquad(n=106)$$ ($$dz=\color{red}{-}dw/w^2$$ is compensated by the change of orientation of the contour).

Hence, it is $$2\pi\mathrm{i}f'(0)$$ where $$f(w)$$ is the cosine above. And the result is clearly zero.