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I want to calculate $\mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+a^2)}\right\}$ using the convolution theorem $\mathcal{L}\{f*g\}=\mathcal{L}\{f\}\cdot\mathcal {L}\{g\}$. I have already calculated it using partial fraction decomposition which yielded $\frac{t}{a^2} - \frac{\sin(at)}{a^3}$.

My approach:

$$f(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t$$

$$g(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2+a^2}\right\} = \frac{1}{a}\sin(at)$$

$$\mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+a^2)}\right\} = f*g = \int_{-\infty}^{\infty} f(t-\tau)g(\tau)\,\mathrm{d}\tau = \frac{1}{a}\int_{-\infty}^{\infty} (t-\tau)\sin(a\tau)\,\mathrm{d}\tau$$

but the last integral is clearly divergent. Where did I go wrong?

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    $\begingroup$ You forgot the unit step functions on $f$ and $g$. Both functions are $0$ for $t<0$ So, the upper limit of the convolution integral is $t$. $\endgroup$
    – Mark Viola
    Jul 3, 2018 at 16:18
  • $\begingroup$ @MarkViola Thanks, this solves the problem. Want to make this an answer so I can accept it? $\endgroup$
    – Fytch
    Jul 3, 2018 at 16:25
  • $\begingroup$ You're welcome. My pleasure. I've posted a solution as you suggested. $\endgroup$
    – Mark Viola
    Jul 3, 2018 at 19:27

1 Answer 1

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Note that we have

$$f(t)=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}=tu(t)$$

and

$$g(t)=\mathscr{L}^{-1}\left\{\frac1{s^2+a^2}\right\}=\frac{\sin(|a|t)}{|a|}u(t)$$

Then, application of the convolution theorem yields

$$\begin{align} \mathscr{L}^{-1}\left\{\frac{1}{s^2(s^2+a^2)} \right\}&=(f*g)(t)\\\\ &=\int_{-\infty}^\infty f(t-\tau)g(\tau)\,d\tau\\\\ &=\int_{-\infty}^\infty (t-\tau)u(t-\tau)\frac{\sin(|a|\tau)}{|a|}u(\tau)\,d\tau\\\\ &=\int_0^t (t-\tau)\frac{\sin(|a|\tau)}{|a|}\,d\tau\tag1 \end{align}$$

We leave it as an exercise to evaluate $(1)$.

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