# Finding limits of integration in convolution

I am struggling to fully get how to choose proper limits of integration when calculating convolutions. Right now I am stuck on a problem where I have to show that when taking the Fourier transform of the convolution of two rectangular functions, the result equals the square sinc function.

So, I have defined:

$f(t) = 1$ for $-1 < t < 1$

$f(t) = 0$ otherwise

By taking the convolution of this function with itself, I have set up:

$$f \ast f (t)= \int_{- \infty}^{\infty} f(\tau) f(t - \tau) d \tau$$

And this is where I struggle to find the correct limits of integration. I know that when I solve this, it is an easy matter to set up a Fourier transform, and since this transform is integrated with respect to $t$, it is easy to see that the limits of integration then will be $-1$ and $1$. But how do I determine the limits here when we integrate with respect to $\tau$? Any help will be greatly appreciated!

• You will never find sinc here. Commented Feb 8, 2013 at 12:52
• I know that I will not find sinc^2 by just performing the convolution. I have to also take the FT. Or is it some other reason for that? Anyways, my main question here is really about how to choose proper limits of integration. Commented Feb 8, 2013 at 13:06

First, since $f(\tau)=0$ outside of $[-1,1]$, we have $$f\ast f(t)=\int_{-1}^1f(t-\tau)d\tau.$$

Now change the variable by $u=t-\tau$: $$f\ast f(t)=\int_{t-1}^{t+1}f(u)du.$$

If $t+1\leq -1$ (ie $t \leq -2$), or if $t-1\geq 1$ (ie $t\geq 2$), then $(-1,1)\cap (t-1,t+1)=\emptyset$ so $f\ast f(t)=0$.

Let us consider the case $-2<t<2$ now.

First, if $-2<t\leq 0$, then $1+t\leq -1$ and $[t-1,t+1]\cap [-1,1]=[-1,t+1]$ so $$f\ast f(t)=\int_{-1}^{t+1}1du=2+t.$$

Second, if $0\leq t<2$ then $1-t\geq 1$ and $[t-1,t+1]\cap [-1,1]=[t-1,1]$ so $$f\ast f(t)=\int_{t-1}^{1}1du=2-t.$$

• Great! Thanks a lot! Commented Feb 8, 2013 at 13:51
• Hey, you're welcome! Commented Feb 8, 2013 at 13:55
• Hi, if I have $-2 \le t \le 0$, I add +1 to both sides to get $-1 \le t+1 \le 1$. Why do you say then that $1+t \le -1$? Commented Jun 13, 2014 at 4:03