$\int_{C^+(0,R)} \frac {dz} {(z^2-1)\dots(z^2-100)}$ independent of $R$ if $R>10$ How can I show that $$\int\limits_{C^+(0,R)} \frac {dz} {(z^2-1)\dots(z^2-100)}$$ is independent of $R$ if $R>10$ without calculating all the residues?
 A: The answer is simple: because when $R>10$, the contour encloses all of the poles of the integrand.  As $R$ increases, the poles and their residues remain the same.  Thus, the value of the integral remains the same.
A: An idea: since all the poles of the integrand function are simple and appear in pairs of opposite sign, show their residue cancel each other:
$$\forall\,1\le k\le 100\;,\;k\in\Bbb N\;\;,\;\;f(z):=\frac{1}{(z^2-1)\cdot\ldots\cdot (z^2-k)\cdot\ldots\cdot(z^2-100)}$$
then
$$Res_{z=\sqrt k}(f)=\lim_{z\to\sqrt k}(z-\sqrt k)f(z)\stackrel{l'Hospital,\,say}=\frac{1}{(k^2-1)\cdot\ldots\cdot 2\sqrt k\cdot\ldots\cdot (k^2-100)}$$
$$Res_{z=-\sqrt k}(f)=\lim_{z\to-\sqrt k}(z+\sqrt k)f(z)\stackrel{l'Hospital,\,say}=\frac{1}{(k^2-1)\cdot\ldots\cdot (-2\sqrt k)\cdot\ldots\cdot (k^2-100)}$$
So there! The residues cancel each other in the pairs of opposite poles and we're done (of course, the important part is that $\,R>10\Longrightarrow\,$ all the poles are within the domain inclosed by the circle.)
