# Integral of Dirac-delta function from convolution theorem

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?

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 $$\omega(t)=\boldsymbol{1}_{(0,t)}(b)\cos(a(t-b))$$
• does $1_{(0,t)}$ just mean the $1$ is only true for $b\in(0,t)$? Commented Apr 25, 2019 at 18:29
• @HenryLee It supposed to be the indicator function of $(0, t)$, i.e. it's $1$ if $b \in (0, t)$ and $0$ otherwise. I think you can use the unit step function as well. Commented Apr 25, 2019 at 18:30