I have been trying to calculate the convolution $t \ast t^2 \theta(t)$ where $\theta(t)$ is the Heaviside theta like so: $$\int_{-\infty}^{+\infty} (t-\tau)\tau^2\theta(\tau)d\tau = \int_0^{+\infty} (t-\tau)\tau^2d\tau$$ The problem is that I can't understand why does it even converge (Mathematica says that $t \ast t^2 \theta(t) = \frac{t^4}{12}$). Answer looks like the integral should have $t$ as the upper limit but it doesn't seem right in this case.
I've also tried to use Fourier transform but couldn't apply inverse since the image is $-2\frac{\delta'(w)}{w^3}+i\pi \delta'(w)\delta''(w)$
It is easy however to calculate $t \theta(t) \ast t^2 \theta(t) = \frac{t^4}{12}\theta(t)$. But how do I get the answer for the original convolution?
EDIT: Thanks for the responses, it is clear now that this convolution does not exist. So I guess, there is a bug in Mathematica as well :)
The upper limit should be $t$ \begin{eqnarray*} \int_0^{\color{red}{t}} (t-\tau)\tau^2d\tau = \cdots \end{eqnarray*}