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Subsection 6.3.2 Laplace Transform of the Derivative of \(f(t)\)

If \(\mathscr{L}[f(t)] =F(s)\) then
\begin{equation*} \mathscr{L}[f'(t)] = sL[f(t)]-f(0) \end{equation*}

Proof.

we have
\begin{equation*} \mathscr{L}[f'(t)]=\int\limits_{0}^{\infty}e^{-st}f'(t)\,dt \end{equation*}
\begin{equation*} = \left.e^{-st}f(t)\right\vert_{0}^{\infty}-\int\limits_{0}^{\infty}-se^{-st}f(t)\,dt \end{equation*}
\begin{equation*} =0-f(0)+s\int\limits_{0}^{\infty}e^{-st}f(t)\,dt = sL[f(t)]-f(0) \end{equation*}

Subsubsection 6.3.2.1 Laplace Transform of Derivative of order \(n\)

\begin{equation*} \mathscr{L}[f^{n}(t)] =s^{n}\mathscr{L}[f(t)]-s^{n-1}f(0)-s^{n-2}f' (0)-s^{n-3}f''(0) -\cdots-f^{n-1}(0). \end{equation*}

Proof.

we have,
\begin{equation} L[f'(t)]=sL[f(t)]-f(0)\tag{6.3.1} \end{equation}
replacing \(f(t)\) by \(f'(t)\) and \(f'(t)\) by \(f''(t)\) in equation (6.3.1), we get -
\begin{equation*} L[f''(t)]=sL[f'(t)]-f'(0) = s\{sL[f(t)]-f(0)\}-f'(0) \end{equation*}
\begin{equation*} =s^{2}L[f(t)]-sf(0)-f'(0) \end{equation*}
Similarly,
\begin{equation*} L[f'''(t)]=s^{3}L[f(t)]-s^{2}f(0)-sf'(0)-f''(0) \end{equation*}
\begin{equation*} \therefore L[f^{n}(t)] =s^{n}L[f(t)]-s^{n-1}f(0)-s^{n-2}f' (0)-s^{n-3}f''(0) -\cdots-f^{n-1}(0). \end{equation*}

Subsubsection 6.3.2.2 Laplace Transform of Integral of \(f(t)\)

If \(\mathscr{L}[f(t)] =F(s)\text{,}\) then
\begin{equation*} \mathscr{L}\left[\int\limits_{0}^{t}f(t)\,dt\right]=\frac{1}{s}F(s). \end{equation*}

Proof.

Let
\begin{equation*} \phi (t) = \int\limits_{0}^{t}f(t)\,dt \end{equation*}
then
\begin{equation*} \phi' (t) = \frac{\,d}{\,dt}\left[\int\limits_{0}^{t}f(t)\,dt\right] =f(t) \end{equation*}
and
\begin{equation*} \phi (0) = \int\limits_{0}^{0}f(t)\,dt=0 \end{equation*}
Now
\begin{equation*} \mathscr{L}[\phi' (t)]=s\mathscr{L}[\phi (t)]-\phi (0)=s\mathscr{L}[\phi (t)] \end{equation*}
or,
\begin{equation*} \mathscr{L}[\phi (t)]=\frac{1}{s}\mathscr{L}[\phi' (t)] \end{equation*}
or,
\begin{equation*} \mathscr{L}\left[\int\limits_{0}^{t}f(t)\,dt\right] = \frac{1}{s}\mathscr{L}[f(t)]=\frac{1}{s}F(s) \end{equation*}