why does the circle at infinity not contribute the integral? I am reading an online physics paper consider the integral
$$\oint_C\frac{ds}{e^{2\pi i s}-1}\frac{e^{isu}}{s^2(s^2-1)} $$
where the contour $C$ is the union of a big counter clockwise circle at infinity and a clockwise contour circling the poles at $s=0,\pm1$.Suppose $0<u<2\pi$. Then the author states that the circle at infinity does not contribute.  Who can explain it in detail ? Thanks in advance.
Renew the background 
The author originally wants to calculate the series
$$\sum_{n \neq 0,\pm1}\frac{e^{inu}}{n^2(n^2-1)}$$
Then he changes the sum to the contour integral
$$\oint_{C'}\frac{ds}{e^{2\pi i s}-1}\frac{e^{isu}}{s^2(s^2-1)}$$
where the contour ${C'}$ is the small circles circling $s=\pm2,\pm3,\cdots$. And then the contour $C'$ can be deformed to the contour $C$ which is is the union of a big counter clockwise circle at infinity and a clockwise contour circling the poles at $s=0,\pm1$.
 A: The contour integral above is equal to
$$i R_N \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{e^{i u R_n e^{i \theta}}}{e^{i 2 \pi R_N e^{i \theta}}-1} \frac1{R_N^2 e^{i 2 \theta} \left ( R_N^2 e^{i 2 \theta}-1 \right ) } \\ - i r_1 \int_0^{2 \pi} d\phi \, e^{i \phi} \frac{e^{i u r_1 e^{i \phi}}}{e^{i 2 \pi r_1 e^{i \phi}}-1} \frac1{r_1^2 e^{i 2 \phi} \left ( r_1^2 e^{i 2 \phi}-1 \right ) } $$
where $R_N=N+1/2$, $N \gt 1$ and $N \in \mathbb{Z}$ and $1 \lt r_1 \lt 2$. That is, the first integral represents the outer circle and the second integral represents the inner circle.  It is straightforward to show that the first integral approaches zero as $N \to \infty$.  For example, consider the magnitude of the integral, which is bounded by the maximum of the integrand times the length of the integration interval.  The magnitude of the integrand is bounded by
$$\frac1{R_N^3} \frac{e^{-u R_N \sin{\theta}}}{\sqrt{e^{-4 \pi R_N \sin{\theta}} - 2 e^{-2 \pi R_N \sin{\theta}} \cos{(2 \pi R_N \cos{\theta})} + 1}} \left | 1-\frac1{R_N^2 e^{i 2 \theta}} \right |^{-1}  $$
The magnitude is a maxima when $\theta=0$.  Because $R_N=N+1/2$, the cosine factor in the denominator is simply $-1$, so the maximum of the integrand in the limit as $N \to \infty$ is simply $1/2$.  Thus, the integral vanishes as $ \pi/(R_N^3)$ as $N \to \infty$.
