I am trying to prove that $$ \lim_{r\to 0}\int_{\pi}^{2\pi} \frac{1 - \exp\left(2\mathrm{i}r\mathrm{e}^{\mathrm{i}t}\right)}{4r^{2}\mathrm{e}^{\mathrm{i}t}}\,\mathrm{i}\,\mathrm{d}t = \frac{\pi}{2} $$
Since I don't know how to solve the integral explicitly, I don't know what to do. This problem is really similar to this one, maybe that answer is useful for this problem too.
Thank you.
On
Since the problem that you linked discussed the limit as $r\rightarrow \infty,$ I'll say something about that, in case that is what you mean. I'll make the same assumption that the other response did regarding your definition of the function.
Call the integrate $f(z)=\frac{1-e^{2iz}}{4z^2}.$ Note that it has a simple pole at $z=0$. Then, calculating the residue at $z=0$ gives $Res_{z=0}(f)=-\frac{i}{2}.$ Applying the residue theorem to the disk $D_r(0)$, we have that $$\int\limits_{\partial D_r(0)}\frac{1-e^{2iz}}{4z^2}dz=2\pi i\left(\frac{-i}{2}\right)=\pi.$$ Note that $$\int\limits_{\partial D_r(0)}\frac{1-e^{2iz}}{4z^2}dz=\int\limits_{\Gamma_1}\frac{1-e^{2iz}}{4z^2}dz+\int\limits_{\Gamma_2}\frac{1-e^{2iz}}{4z^2}dz,$$ where $\Gamma_1$ is the contour in the linked problem and $\Gamma_2$ is your contour. Taking the limit as $r\rightarrow\infty,$ the integral along $\Gamma_1$ vanishes, yielding that your integral is equal to $\pi.$
If you are integrating $\int_C f(z)\,\mathrm{d}z$ for $f(z)=(1-e^{2iz})/z^2$ with the parametrization $\require{color}C\colon z=re^{it}$, you forgot an $r$ in $\mathrm{d}z={\color{red}r}ie^{it}\,\mathrm{d}t$.
For $0<2r<1$ and $t\in\mathbb{R}$, we have $\lvert 2ire^{it}\rvert=2r<1$ and hence $$ \frac{1-e^{2ire^{it}}}{4r^2e^{it}}=\frac{1-(1+2ire^{it}+o(r))}{4r^2e^{it}}=\frac{-2ire^{it}+o(r)}{4r^2e^{it}}=-\frac{i}{2r}+o(r^{-1}) $$ Now integrate, $$ \int_{\pi}^{2\pi}\frac{1-e^{2ire^{it}}}{4r^2e^{it}}{\color{red}r}i\,\mathrm{d}t= \int_{\pi}^{2\pi}\left[-\frac{i}{2r}+o(r^{-1})\right]{\color{red}r}i\,\mathrm{d}t=\frac\pi2+\underbrace{o(1)}_{\to 0}. $$