# Convolution integral of $t$ and $t^2\theta(t)$

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 convolution of two generalized functions with unbounded supports is defined provided that either both supports are bounded on the left or both supports are bounded on the right. This one is undefined. Jan 14, 2020 at 14:06

The upper limit should be $$t$$ $$\begin{eqnarray*} \int_0^{\color{red}{t}} (t-\tau)\tau^2d\tau = \cdots \end{eqnarray*}$$
• Thank you but could you please clarify why? Since $f(t)=t$ is supported on $(-\infty; +\infty)$ I just can't get why there shouldn't be $+\infty$ in the upper limit. Jan 13, 2020 at 0:48
• The first function needs to be $t \theta(t)$. Jan 13, 2020 at 1:03
• Well, yes, if it is indeed $t\theta(t)$ the solution is clear. But the original exercise is to convolve specifically $f(t)=t$ and I would accept your answer had Wolfram Mathematica not given $\frac{t^4}{12}$ as the answer. Maybe it's just wrong, I don't know :) Jan 13, 2020 at 1:11