Inverse Laplace transform of a product using convolution 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?

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 $(1)$.
