Answer

Given

\[ \begin{aligned} f(t) &= e^{t}\roundbr{u(t) - u(t - \pi/2)}\newline L(f) &= L\roundbr{e^{t - 0}u^{t - 0}} - L\roundbr{e^{t}u\roundbr{t - \frac{\pi}{2}}}\newline &= e^{0s}\frac{1}{s-1} - e^{-\frac{\pi}{2}s}L\roundbr{e^{t + \pi/2}}\newline &= \frac{1}{s-1} - e^{-\frac{\pi}{2}s}e^{\pi/2}L\roundbr{e^{t}}\newline &= \frac{1}{s-1} - e^{-\frac{\pi}{2}s}e^{\pi/2}\frac{1}{s-1}\newline &= \frac{1 - e^{-\frac{\pi}{2} (s-1)}}{s-1} \end{aligned} \]

using two different versions of the second shift theorem.