TITLE: Evaluate the integration $\int_{|z|=100}f(z)\,dz$
QUESTION [5 upvotes]: Let , $$f(z)=\frac{1}{z}.\frac{1-2z}{z-2}\cdots \frac{1-10z}{z-10}$$Find $$\int_{|z|=100}f(z)\,dz$$

We find that the function $f(z)$ has simple pole at the points $z=0,2,4,6,8,10$ , and all the points lie in $|z|=100$. So the required integral equal to $2\pi i\times\text{sum of the residues}$. But that process is too much laborious in this case.
Does there any other simplest way to evaluate the integral ?

REPLY [3 votes]: Inasmuch as all of the singularities are contained within $|z|<100$, we can evaluate the contour integral by finding the Residue at Infinity.  To do this, we evaluate 
$$\text{Res}\left(f(z),z=\infty\right)=\text{Res}\left(-\frac{1}{z^2}f\left(\frac1z\right),z=0\right)$$
Now, given $f(z)=\frac1z\prod_{n=2,n even}^{10}\frac{1-nz}{z-n}$, we find that 
$$\bbox[5px,border:2px solid #C0A000]{f\left(\frac1z\right)=z\prod_{n=2,n even}^{10}\frac{z-n}{1-nz}}$$
Therefore, 
$$\text{Res}\left(f(z),z=\infty\right)=-\lim_{z\to 0}\left(\prod_{n=2,n even}^{10}\frac{z-n}{1-nz}\right)=\prod_{n=2,n even}^{10}\,n=3840$$
Finally, we have 
$$\bbox[5px,border:2px solid #C0A000]{\oint_{|z|=100} f(z)\,dz=-2\pi i (3840)}$$