Question

Can someone please check this; \[\begin{equation*}
f(t)=\cos(t)\big(h(t-\pi)-h(t)\big)
\end{equation*}\]

\[
\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal
\newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x

\begin{align*}
\intl0\infty{f(t)e^{-pt}}t
\end{align*}
\]

Answer 1

\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal
\newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x
\begin{align*}
\intl0\infty{f(t)e^{-pt}}t&=\intl0\infty{\cos(t)\big(h(t-\pi)-h(t)\big)e^{-pt}}t\\
&=\intl0\infty{\cos(t)h(t-\pi)e^{-pt}}t-\intl0\infty{\cos(t)h(t)e^{-pt}}t\\
&=\intl\pi\infty{\cos(t)e^{-pt}}t-\intl0\infty{\cos(t)e^{-pt}}t\\
&=\left.\frac{e^{-pt}\big(-p\cos t+\sin t\big)}{(-p)^2+1^2}\right|_\pi^\infty-\left.\frac{e^{-pt}\big(-p\cos t+\sin t\big)}{(-p)^2+1^2}\right|_0^\infty\\
&=\frac{-e^{-p\pi}\big(-p\cos \pi+\sin \pi\big)}{p^2+1}+\frac{\big(-p\cos 0+\sin 0\big)}{p^2+1}\\
&=\frac{-pe^{-p\pi}}{p^2+1}-\frac{p}{p^2+1}\\
\\
&=\frac{-p\big(1+e^{-p\pi }\big)}{p^2+1}\\
\end{align*}\]

Answer 2

I haven't had my coffee yet, but I don't quite see how you managed to make h(t) vanish

Answer 3

I think h is the step function

Answer 4

ah, okay, sure, I didn't realize that we could assume anything about h(t)

Answer 5

isn't that Laplace transform? .. let p = s

Answer 6

The unit heaviside step function \[h(t-a) =\begin{cases}0&t<a\\ 1&t≥a\end{cases} \]