Question

\[
\mathcal L ^{-1}\left\{\frac{e^{-\pi p}}{p^2+1}\right\}\]

Answer 1

\[ \begin{align*}
\mathcal L ^{-1}\left\{\frac{e^{-\pi p}}{p^2+1}\right\}
&=h(t-\pi)\times\mathcal L^{-1} \left\{\frac1{p^2+1}\right\}_{t\rightarrow t-\pi}\\
&=h(t-\pi)\times\sin(t)\Big|_{t\rightarrow t-\pi}\\
&=h(t-\pi)\times\sin(t-\pi)\\
&=-h(t-\pi)\sin(t)\\
\end{align*}\]

Answer 2

@AccessDenied

Answer 3

have i got this one right/?

Answer 4

This appears correct to me. :)

Answer 5

thanks!,

would you more likely leave the answer in the final form of
\[=h(t−π)\sin(t−π)\]or
\[=−h(t−π)\sin(t)\]

Answer 6

I would probably use -h(t-pi) sin(t) just because I prefer to see the simplest forms of functions..