Here is my function $f(z)=\frac{1}{(z+1)(z+2)(z+3)...(z+100)}$ I need to find this $\int f(z)dz$ where $C:|z|=150$ counterclockwise
I tried to use second residue theorem here. So i get $\int f(z)dz=2\pi i Res_{z=0}[1/z^2 f(1/z)$ so my function here: $f(1/z)=\frac{1}{(1/z+1)(1/z+2)(1/z+3)...(z+100)}$ $1/z^2 f(1/z)=1/z^2((1/z+1)(1/z+2)(1/z+3)...(1/z+100))$
But after that i tried to use $Res_{z=0}[1/z^2 f(1/z)=lim_{z\to 0}\frac{1}{1/z^2(z+1)(z+2)(z+3)...(z+100)}$ So i get 100 times $1/0$ and that not make sense
So any what should i do to solve this integral?
Since $f(1/z)=\prod_{j=1}^{100}\tfrac{1}{1/z+j}=\prod_j\frac{z}{z+j}$, the correct calculation is$$z^{-2}f(1/z)=\tfrac{z^{98}}{\prod_j(z+j)}\stackrel{z\to0}{\to}0,$$making the integral $0$. As a sanity check, compare with $R>2$ to$$\oint_{|z|=R}\tfrac{dz}{(z+1)(z+2)}=\oint(\tfrac{1}{z+1}-\tfrac{1}{z+2})dz=2\pi i-2\pi i=0.$$