Evaluate the given limit in $C_r=\{re^{i\theta}:0\le \theta \le \pi\}$ 
Let , $C_r=\{re^{i\theta}:0\le \theta \le \pi\}$ denotes the semicircle traversed clockwise. Show that $$\lim_{r\to 0}\int_{C_r}\frac{e^{iz}}{z(z^2+1)}\,dz=-\pi i$$

I can not use the Jordan's theorem, as the limit $r\to 0$. I tried through putting $z=re^{i\theta}$ , but it does not help..How I find the limit ?
 A: It is not difficult to show that if $f(z)$ is a holomorphic function in a neighbourhood of the origin we have:
$$ \lim_{r\to 0}\int_{C_r}\frac{f(z)\,dz}{z^2+1} = 0 $$
since both $|f(z)|$ and $\left|\frac{1}{z^2+1}\right|$ are bounded in a neighbourhood of the origin, while the length of the integration path is $\pi r\to 0$. $\frac{e^{iz}-1}{z}$ is an entire function, hence:
$$ \lim_{r\to 0}\int_{C_r}\frac{e^{iz}}{z(z^2+1)}\,dz = \lim_{r\to 0}\int_{C_r}\frac{dz}{z(z^2+1)}=\lim_{r\to 0}\int_{C_r}\frac{dz}{z}\tag{1} $$
where the last equality follows from the same argument, since $$\int_{C_r}\left(\frac{1}{z(z^2+1)}-\frac{1}{z}\right)\,dz =-\int_{C_r}\frac{z\,dz}{z^2+1}$$ is the integral of a holomorphic function in a neighbourhood of the origin.
The last integral in $(1)$ can be computed in a explicit way by substituting $z=r e^{i\theta}$:
$$ \int_{C_r}\frac{dz}{z} = -ri\int_{0}^{\pi}\frac{d\theta}{r} = -\pi i.\tag{2}$$
A: $$\begin{align}
\lim_{r\to 0}\int_{C_r}\frac{e^{iz}}{z(z^2+1)}\,dz&=-\lim_{r\to 0}\int_0^{\pi} \frac{e^{ire^{i\phi}}} {re^{i\phi}(r^2e^{i2\phi}+1)}ire^{i\phi}d\phi\\\\
&=-\lim_{r\to 0}\int_0^{\pi} \frac{ie^{ire^{i\phi}}} {r^2e^{i2\phi}+1}d\phi \tag 1\\\\
&=-\int_0^{\pi} \lim_{r\to 0}\left(\frac{ie^{ire^{i\phi}}} {r^2e^{i2\phi}+1}\right)d\phi\tag 2\\\\
&=-\int_0^{\pi}\,i\,d\phi\\\\
&=-\pi i
\end{align}$$

NOTE:
In going from $(1)$ to $(2)$ the interchange of the limit and the integral is justified by the Dominated Convergence Theorem.  Here, 
$$\left|\frac{ie^{ire^{i\phi}}} {r^2e^{i2\phi}+1}\right| \le \frac{1}{|1-r|^2}$$
and obviously the constant $\frac{1}{|1-r|^2}$, $r\ne 1$, is integrable on the interval of interest.
A: More generally, given a function $f$ with a simple pole at $z= \alpha$, we have:
$$
\lim_{\epsilon\rightarrow 0}\int_{\Delta_{\epsilon,\theta}}{f(z)dz}=i\theta\cdot \text{Res}(f,\alpha). 
$$
Where $\Delta_{\epsilon,\theta}$ is an arch (with positive direction) about $\alpha$ with radius $\epsilon$, over an angle of size $\theta$.
In your case, set $f(z) = \frac{e^{iz}}{z(z^2+1)}$, $\,\,\alpha=0$ and $\theta=\pi$, and you'll get the result by reversing the direction (multiplying by $-1$).
(Note that in your case: $\text{Res}(f,\alpha) = \lim_{z\rightarrow 0}zf(z)=1$)
Now, lets prove the claim:
Since $\alpha$ is a simple pole, $f$ can be written as follows:
$$
f(z) = \frac{\text{Res}(f,\alpha)}{z-\alpha} + g(z)
$$
Where $g$ is holomorphic in a neighbourhood of $\alpha$.
Let $M$ be an upper bound of $|g|$ in that neighbourhood.
Thus, for small enough $\epsilon$:
$$
\left|\int_{\Delta_{\epsilon,\theta}}{g(z)dz}\right|\le M\cdot\epsilon\theta\tag{1}
$$
Also,
$$
\int_{\Delta_{\epsilon,\theta}}{\frac{dz}{z-\alpha}}=\int_{\theta_0}^{\theta_0+\theta}{\frac{\epsilon i e^{it}}{\epsilon e^{it}}dt}=i\theta\tag{2}
$$
Combining $(1)$ and $(2)$, and setting $\epsilon\rightarrow 0$ we get the result. 
