I''m currently studying Laplace transforms using the textbook Advanced Engineering Mathematics 10e (Kreyszig) and had a question regarding the Laplace transforms of the integral of a function.
The example in the textbook is:
Find the inverse of $\dfrac{1}{s(s^2 + w^2)}$.
Since we know that:
$$
\begin{align}
f(t) & = \sin{(wt)} \\
\mathcal{L}(f) & = \frac{w}{s^2 + w^2}
\end{align}
$$
We also know that:
$$
\mathcal{L}^{-1} \left( \frac{1}{s^2 + w^2} \right) = \frac{1}{w} \sin{(wt)}
$$
What's confusing me is the part where we apply the integral in order to obtain the final solution. The textbook does as follows:
$$
\begin{align}
\mathcal{L}^{-1} \left( \frac{1}{s(s^2 + w^2)} \right) & = \int_0^t \frac{1}{w}\sin{(wt)} d \tau \\
& = \frac{1}{w^2}(1 – \cos{(wt)})
\end{align}
$$
How was the last equation derived? Aren't $t$ and $w$ constants when integrating for $\tau$? Perhaps I'm missing something regarding $\tau$ but there isn't any explanation for it in the textbook.
Thanks in advance.
Best Answer
Let
$$f(t):=\mathcal{L}^{-1}\left( \dfrac{1}{s(s^2+w^2)}(t) \right)$$
so that
$$\dfrac{1}{s(s^2+w^2)}=\int_0^\infty f(t)e^{-st}dt$$ and
$$\dfrac{1}{s^2+w^2}=\int_0^\infty f(t)se^{-st}dt\stackrel{\ast}{=}\int_0^\infty f^\prime(t)e^{-st}dt$$ where $\stackrel{\ast}{=}$ comes from integrating by parts, provided we justify $[f(t)e^{-st}]_0^\infty=0$.
So
$$f^\prime(t) = \mathcal{L}^{-1}\left( \frac{1}{s^2+w^2} \right) \implies f(t) = \int_0^t\mathcal{L}^{-1}\left( \frac{1}{s^2+w^2} \right) (\tau)d\tau$$
provided we also justify $f(0)=0$. The textbook's $wt$ is a misprint for $w\tau$.