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?
On
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 $\theta$. $$ \int_0^{\pi/2}e^{-2r\theta/\pi}\frac{r}{r-1}\,\mathrm{d}\theta +\int_{\pi/2}^\pi e^{-2r(\pi-\theta)/\pi}\frac{r}{r-1}\,\mathrm{d}\theta =\frac\pi{r-1}\left(1-e^{-r}\right) $$
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.