# Laplace transform of integral of a function

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.

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 $$w\tau$$.

• You just proved $\mathcal{L}[f'(t)]=sF(s)-f(0)$, right? Mar 19 at 10:31

There's one thing wrong with one of your hypotheses. You actually have:

$$\mathscr{L}\left( \dfrac{1}{w}\sin(wt) \right)=\dfrac{1}{s^2+w^2}$$

and not the other way around.

The Laplace transform of a convolution of two functions is given by:

$$\mathscr{L}(f \circledast g) = \mathscr{L}(f) \mathscr{L}(g)$$

Make $$f=\dfrac{1}{w}\sin(wt)$$ and $$g=1$$ so that: \begin{align} \dfrac{1}{s(s^2 + w^2)} & = \mathscr{L}\left( \dfrac{1}{w} \sin(wt) \right) \mathscr{L}(1) \\ & =\mathscr{L}\left(1 \circledast \dfrac{1}{w}{\sin(wt)} \right) \\ & = \int_0^t \dfrac{1}{w}\sin(wx) 1(1-x) dx \\ & = \dfrac {1}{w^2}(1-\cos(wt)) \end{align}

• @Sean, your proposition is really correct, but the step in the convolution with $1(t)$ is incorrect! $\mathscr{L}\left(1 \circledast \dfrac{1}{w}{\sin(wt)} \right)=\int_0^t \dfrac{1}{w}\sin(wx)dx=\dfrac {1}{w^2}(1-\cos(wt))$. Mar 19 at 10:46

I guess , they have applied Convolution theorem Here it is https://en.m.wikipedia.org/wiki/Convolution_theorem

Using the fact that Laplace of 1 is $$1/s$$

• This is quite strange though... Convolutions are in the same chapter but at a much later section.
– Sean
Nov 29, 2019 at 6:52
• The way they have written the solution , I don't think they are applying something other than Convolution Nov 29, 2019 at 7:19
• Yes, I agree I don't see any other way that they achieved this result. However, I really don't think it would be appropriate of the authors to suddenly use convolution operations without even having taught them yet. :(
– Sean
Nov 29, 2019 at 8:12
• Yeah I agree with you . Nov 29, 2019 at 8:14