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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.

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3 Answers 3

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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 $w\tau$.

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  • $\begingroup$ You just proved $\mathcal{L}[f'(t)]=sF(s)-f(0)$, right? $\endgroup$
    – MathArt
    Mar 19 at 10:31
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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: $$ \begin{align} \dfrac{1}{s(s^2 + w^2)} & = \mathscr{L}\left( \dfrac{1}{w} \sin(wt) \right) \mathscr{L}(1) \\ & =\mathscr{L}\left(1 \circledast \dfrac{1}{w}{\sin(wt)} \right) \\ & = \int_0^t \dfrac{1}{w}\sin(wx) 1(1-x) dx \\ & = \dfrac {1}{w^2}(1-\cos(wt)) \end{align} $$

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

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  • $\begingroup$ This is quite strange though... Convolutions are in the same chapter but at a much later section. $\endgroup$
    – Sean
    Nov 29, 2019 at 6:52
  • $\begingroup$ The way they have written the solution , I don't think they are applying something other than Convolution $\endgroup$
    – Om Prakash
    Nov 29, 2019 at 7:19
  • $\begingroup$ 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. :( $\endgroup$
    – Sean
    Nov 29, 2019 at 8:12
  • $\begingroup$ Yeah I agree with you . $\endgroup$
    – Om Prakash
    Nov 29, 2019 at 8:14

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