\begin{align*}
\int_0^{\frac{\pi}{2}}\exp (t+it)\,dt \amp = \int_0^{\frac{\pi}{2}}e^te^{it}dt\\
\amp = \int_0^{\frac{\pi}{2}}e^t(\cos t+i\sin t \,dt\\
\amp = \int_0^{\frac{\pi}{2}}e^t\cos t \, dt + i\int_0^{\frac{\pi}{2}}e^t\sin t \,dt\text{.}
\end{align*}
We can evaluate each of the integrals via integration by parts. For example,
\begin{align*}
\int_0^{\frac{\pi}{2}}\underset{u}{\underbrace{e^t}} \; \underset{dv}{\underbrace{\cos t\,dt}} \amp = (\underset{u}{\underbrace{e^t}} \; \underset{v}{\underbrace{\sin t}}) \Big|_{t=0}^{t=\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\underset{v}{\underbrace{\sin t}} \; \underset{du}{\underbrace{e^tdt}}\\
\amp = (e^{\frac{\pi}{2}}\sin \frac{\pi}{2}-e^0\sin 0) -\int_0^{\frac{\pi}{2}}\underset{u}{\underbrace{e^t}} \; \underset{dv}{\underbrace{\sin t \,dt}}\\
\amp = e^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\underset{u}{\underbrace{e^t}} \; \underset{dv}{\underbrace{\sin t \,dt}}\\
\amp = e^{\frac{\pi}{2}} - (\underset{u}{\underbrace{e^t}} \; [\underset{v}{\underbrace{-\cos t}}] ) \Big|_{t=0}^{t=\frac{\pi}{2}} +\int_0^{\frac{\pi}{2}}\underset{v}{\underbrace{-\cos t}} \; \underset{du}{\underbrace{e^t\,dt}}\\
\amp = e^{\frac{\pi}{2}}-1-\int_0^{\frac{\pi}{2}}e^t\cos t\,dt\text{.}
\end{align*}
Adding \(\int_0^{\frac{\pi}{2}}e^t\cos t\,dt\) to both sides of this equation and then dividing by 2 gives \(\int_0^{\frac{\pi}{2}}e^t\cos t)\,dt=\frac{1}{2}(e^{\frac{\pi}{2}}-1)\text{.}\) A similar computation procedure yields \(i\int_0^{\frac{\pi}{2}}e^t\sin t\,dt=\frac{i}{2}(e^{\frac{\pi}{2}}+1)\text{.}\) Therefore,
\begin{equation*}
\int_0^{\frac{\pi}{2}}\exp (t+it)\,dt=\frac{1}{2}(e^{\frac{\pi}{2}}-1) + \frac{i}{2}(e^{\frac{\pi}{2}}+1)
\end{equation*}