I need help on the following line integral: $$\int\limits_\gamma f dz = \int\limits_\gamma \frac{1}{g} dz = \int\limits_\gamma \frac{1}{(z + 1)(z + 2) \cdot \ldots \cdot (z + r)} dz$$ For a fix $r \in \mathbb{N} : r \ge 1$ where $\gamma : [0, 1] \to \mathbb{C}, \gamma(t) = 2r \cdot e^{2\pi i t}$. So, the line is the border of a disk with radius $2r$, i.e. $\partial D_{2r}(0)$ if you are more familiar with this notation.
I am allowed to use the full tool box of the complex analysis for this exercise.
We directly see all singularities of the function: $-1, -2, \ldots , -r$.
Also they are isolated, and obviously all of them are simple poles. The winding number of all is $1$.
The integral is given by: $$\int\limits_\gamma fdz = 2\pi i \cdot (\operatorname{res}_{-1} + \operatorname{res}_{-2} + \ldots + \operatorname{res}_{-r})$$
So far so good. Now I know some methods for computing residue of simple poles, for example:
$$\operatorname{res}_{-a} = (z + a) f(z)|_{z = -a} = \frac{1}{(1 - a)(2 - a) \cdot \ldots \cdot ((a - 1) - a) \cdot ((a + 1) - a) \cdot \ldots \cdot (r - a)}$$
Or this one:
$$\operatorname{res}_{-a} = \frac{1}{g'(-a)}$$
However, those approaches seem to be very complex as I do not see where some term simplifies. I assume something simplifies when everything is put together in $(\operatorname{res}_{-1} + \operatorname{res}_{-2} + \ldots + \operatorname{res}_{-r})$ but I can not see the trick.
Thanks for your help :)
If we assume $r\geq 2$ and we consider the integral over a large circle centered at the origin with radius $R$, such integral $\phantom{}^{(A)}$ is bounded by a quantity tends to zero as $R\to +\infty$, so the sum of the residues (at $z=-1,z=-2,\ldots,z=-r$) is zero, just like the integral over $|z|=2r$. The case $r=1$ is trivial.
$(A)$: For instance, in the $r=3$ case: $$ \left|\int_{|z|=R}\frac{dz}{(z+1)(z+2)(z+3)}\right|\leq \frac{2\pi R}{(R-3)^3}.$$