Calculate the convolution of $p_2(x)*u(x)$

Question

$$p_2(x)=\begin{cases} 1 \qquad x \in [-1,1] \\ 0 \qquad x \notin [-1,1] \end{cases}$$

$$u(x)=\begin{cases} 1 \qquad x \ge 0 \\ 0 \qquad x < 0 \end{cases}$$

First, I have calculated the Fourier transforms of $u(x)$ and $p_2(x)$

$$U(\omega)=\int _{0}^{\infty} e^{-i\omega x} dx=-\frac{i}{\omega} $$

$$P_2(\omega)=\int _{-1}^{1} e^{-i\omega x} dx=2 \frac{\sin(\omega)}{\omega}$$

Then, $$U(\omega) \ P_2(\omega)=-2i \frac{\sin(\omega)}{\omega^2}$$

How can I calculate the inverse Fourier transform to calculate $p_2(x)*u(x)$ ?

Thanks!

Answer 1

Why Fourier transformation ? We have

$p_2(x)*u(x)=\int_{-1}^1u(x-s)ds$.

The last integral is easy to compute !