In a question I have been lead to use the convolution theorem to find the inverse Laplace transform, as shown below:
$$\omega(t)=\mathscr{L}^{-1}\left[e^{-bs}\frac{s}{s^2+a^2}\right]$$
From the convolution theorem we know that if:
$$\mathscr{L}\left[f(t)\right]=F(s),\,\mathscr{L}\left[g(t)\right]=G(s)$$
then:
$$\mathscr{L}^{-1}\left[F(s)G(s)\right]=g*t=\int_0^tf(\tau)g(t-\tau)d\tau$$
and since:
$$\mathscr{L}^{-1}\left[e^{-bs}\right]=\delta(t-b)$$
$$\mathscr{L}^{-1}\left[\frac{s}{s^2+a^2}\right]=\cos(at)$$
Then we can evaluate the inverse laplace transform as:
$$\omega(t)=\int_0^t\delta(\tau-b)\cos\left[a(t-\tau)\right]d\tau$$
but I am not sure how to evaluate this integral. I initially thought about letting:
$$\cos(t)=\Re\left[e^{it}\right]$$
but this seemed to make it much more complicated. Is integration by parts the best way to go?
Integral of Dirac-delta function from convolution theorem
convolutionintegrationlaplace transform
Related Solutions
Use the fact that $$\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} = \frac{\frac{s+1}{(s+1)^2+1}}{s} = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).$$
In general, we have $$\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .$$
\begin{align*}\mathcal{L}\{f\}({s})\cdot\mathcal{L}\{g\}({s})&=\left(\int_0^\infty e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^\infty e^{-{s}u}{g}(u)\,du\right)\\
&=\lim_{{{L}}\to\infty}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right)
\end{align*}
By Fubini's theorem,
\begin{align*}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right)&=\int_0^{{L}}\int_0^{{L}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du\\
&=\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du,
\end{align*}
where $R_{{L}}$ is the square region $$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}}.$$
Let $T_{{L}}$ be the triangular region
$$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad t+u\leq {{L}}.$$
Provided that ${f}$ and ${g}$ are bounded by exponential functions,
$$\lim_{{{L}}\to\infty}\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\lim_{{{L}}\to\infty}\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du.$$
Now, the function
$$\varphi(v,u)=(v-u,u)$$
maps $D_{{L}}$ bijectively onto $T_{{L}}$, where $D_{{L}}$ is the triangular region
$$0\leq v\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad v\geq u.$$
The component functions of $\varphi$ are
$$t(v,u)=v-u\qquad\mbox{and}\qquad u(v,u)=u,$$
so the Jacobian of $\varphi$ is
$$J\varphi=\det\begin{bmatrix}t_v&t_u\\u_v&u_u\end{bmatrix}=\det\begin{bmatrix}1&-1\\0&1\end{bmatrix}=1.$$
Hence,
$$\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du.$$
By Fubini's theorem,
\begin{align*}&\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du=\int_0^{{L}}\int_0^ve^{-{s}v}{f}(v-u){g}(u)\,du\,dv\\
\\
&=\int_0^{{L}}e^{-{s}v}\int_0^v{f}(v-u){g}(u)\,du\,dv=\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv.
\end{align*}
Hence,
\begin{align*}\lim_{{{L}}\to\infty}\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du&=\lim_{{{L}}\to\infty}\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv\\
\\
&=\int_0^\infty e^{-{s}v}({f}\ast{g})(v)\,dv\\
\\
&=\mathcal{L}\{{f}\ast{g}\}({s}).
\end{align*}
Best Answer
We have $2$ cases: $\tau-b=0$ for some $\tau \in (0, t)$, or not, i.e. wether $b \in (0, t)$ or not. If not, then $\delta(\tau-b)=0$, so the integral is $0$, which means that $\omega(t)=0$ if $b \notin(0, t)$. Otherwise, the dirac delta just "replaces" the $\tau$ in the integral with $b$, so $\omega(t)=\cos(a(t-b))$. Putting it together, we have that $$\omega(t)=\boldsymbol{1}_{(0,t)}(b)\cos(a(t-b))$$