For $r$ a positive real number let $f(r):=\int_{C_{r}} \frac{\sin(z)}{z}dz$, where $C_r$ is the contour $re^{i\theta}$, $0 \leq \theta \leq \pi$. What is $\lim_{r\rightarrow 0} \frac{f(r)}{r}$ ?
I am unable to integrate on the contour since the singularity $0$ lies on the contour.
Any help is appreciated. Thanks in advance.
On
Since,$$(\forall z\in\mathbb C):\frac{\sin z}z=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots,$$the function $\frac{\sin z}z$ has a primitive:$$F(z)=z-\frac{z^3}{3\times3!}+\frac{z^5}{5\times5!}-\cdots$$and therefore\begin{align}\lim_{r\to0}\frac1r\int_{C_r}\frac{\sin z}z\,\mathrm dz&=\lim_{r\to0}\frac{F(re^{\pi i})}r-\frac{F(re^{0\times i})}r\\&=\lim_{r\to0}e^{\pi i}\frac{F(re^{\pi i})}{re^{\pi i}}-\frac{F(re^{0\times i})}r\\&=e^{\pi i}-1\text{ (because $F'(0)=1$)}\\&=-2.\end{align}
$\frac {f(r)} r=i\int_0^{\pi} \frac {sin (re^{it})} r dt$ and $\frac {\sin z }z \to 1$ as $ r \to 0$. Hence the answer is $i\int_0^{\pi}e^{it} dt=-2$.
[ It is easy to justify the interchange of limit and integral in this case].