Question

Find the inverse laplace transform of the arctangent of 1/p

\[\mathcal L^{-1}\{\arctan\tfrac1p\}\]

Answer 1

\[ \newcommand \p \newcommand
\p \Lap [1] {\operatorname{\mathcal L}\left\{\rule{0pt}{2.2ex}#1\right\} }
\p \dd [1] {\,\mathrm d#1 }
\p \de [2] {\frac{\mathrm d #1}{\mathrm d#2} }
\p \Lin [1] {\operatorname{\mathcal L}^{-1}\left\{#1\right\} }
\p \intl [4] {\int\limits_{#1}^{#2}{#3}\dd{#4} }
\begin{align*}
\Lin{\arctan\tfrac1p} &=\Lin{\intl p\infty{F(p)}p} =\frac{f(t)}t\\
\\
\arctan\tfrac1p &=\intl p\infty{F(p)}p \\
\de{}p\left(\arctan\tfrac1p\right) &=\de{}p\intl p\infty{F(p)}p \\
\frac1{1+1/p^2}\cdot-1/p^2 &=F(\infty)\cdot(\infty)'-F(p)\cdot(p)' \\
\frac{-1/p^2}{1+1/p^2} &=-F(p) \\
\frac{1}{p^2+1} &=F(p) \\
\\
\Lin{\frac{1}{p^2+1}} &=\Lin{F(p)} \\
\sin(t) &=f(t) \\
\\
\Lin{\arctan\tfrac1p} &=\frac{\sin t}t
\end{align*} \]