Convergence of a line integral along semi-circular arc There is a line integral in a form,
$$\int_\mathrm{arc} \frac{\exp(iz)}{z^2+1} \, dz$$  
"arc" is a semi-circular line with radius $R$ on the upper half complex plane.
and i know that the integral converges to zero as R goes to infinity.
What about this integral as $R$ goes to infinity?
$$\int_\mathrm{arc} \frac{\exp(iz)}{z+1} \, dz$$ 
I expect that the second integral converges to a fixed constant as $R$ goes to infinity. Am i right? if i am, how can i calculate this constant?
 A: No, I think the second integral also vanishes.  Write the integral as
$$i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta}}}{1+R e^{i \theta}} = i R \int_0^{\pi} d\theta \,  \, e^{i \theta + i R \cos{\theta}} \frac{e^{-R \sin{\theta}}}{1+R e^{i \theta}}$$
The magnitude of the integral is bounded by
$$\frac{2 R}{R-1} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2 R}{R-1} \int_0^{\pi/2} d\theta \, e^{-2  R\theta/\pi} \le \frac{\pi}{R-1} $$
which vanishes as $\pi/R$ as $R \to \infty$.  This is essentially a form of Jordan's lemma.
A: For $0\le\theta\le\frac\pi2$,
$$
\begin{align}
\left|\frac{e^{iz}}{z+1}\right|
&\le\frac{e^{-\mathrm{Im}(z)}}{|z|-1}\\
&\le\frac{e^{-r\sin(\theta)}}{r-1}\\
&\le\frac{e^{-2r\theta/\pi}}{r-1}\\
\end{align}
$$
For $\frac\pi2\le\theta\le\pi$,
$$
\begin{align}
\left|\frac{e^{iz}}{z+1}\right|
&\le\frac{e^{-\mathrm{Im}(z)}}{|z|-1}\\
&\le\frac{e^{-r\sin(\theta)}}{r-1}\\
&\le\frac{e^{-2r(\pi-\theta)/\pi}}{r-1}\\
\end{align}
$$
Multiply by $r$ and integrate in $\theta$.
$$
\int_0^{\pi/2}e^{-2r\theta/\pi}\frac{r}{r-1}\,\mathrm{d}\theta
+\int_{\pi/2}^\pi e^{-2r(\pi-\theta)/\pi}\frac{r}{r-1}\,\mathrm{d}\theta
=\frac\pi{r-1}\left(1-e^{-r}\right)
$$
