I have a doubt about convolution.

I have found this definition :

$$(f*g)(t)=\int_{-\infty}^{+\infty} f(t-\alpha) \ g(\alpha) \ d\alpha$$

This integral does not converge:

$$\cos(t)*t=\int_{-\infty}^{+\infty} \cos(t-\alpha) \ \alpha \ d\alpha$$

Contrariwise: $$ \mathscr{L} \{ \cos(t) * t \} =\mathscr{L} \{ \cos(t) \} \ \mathscr{L} \{t \}=\frac{1}{s^3+s}$$

Partial fraction decomposition:


$$A=\lim_{s\rightarrow 0} \ \frac{1}{s^2+1}=1$$ $$B=\lim_{s\rightarrow i} \ \frac{1}{s^2+is}=-\frac{1}{2}$$ $$C=\lim_{s\rightarrow -i} \ \frac{1}{s^2-is}=\frac{1}{2}$$

$$\frac{1}{s^3+s}=\frac{1}{s}+\frac{-\frac{1}{2}}{s-i}+\frac{\frac{1} {2}}{s+i}$$

$$\mathscr{L}^{-1} \{ \frac{1}{s}+\frac{-\frac{1}{2}}{s-i}+\frac{\frac{1}{2}}{s+i} \}=1-\frac{1}{2} \ e^{it}+\frac{1}{2} \ e^{-it}$$

Then, I have found this other definition of convolution (in a lesson about Laplace transform):

$$(f*g)(t)=\int_{0}^{t} f(t-\alpha) \ g(\alpha) \ d\alpha$$

Why are there two different definitions about convolution?

When have I to use one or another?



3 Answers 3


Let us quote Wikipedia:

The convolution of $f$ and $g$ is [...]

\begin{align} (f * g )(t) & \, \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau) \, d\tau \\ & = \int_{-\infty}^\infty f(t-\tau)\, g(\tau)\, d\tau. \end{align}

For functions $f,g$ supported on only $[0, \infty)$ (i.e., zero for negative arguments), the integration limits can be truncated, resulting in

$$(f * g )(t) = \int_{0}^{t} f(\tau)\, g(t - \tau)\, d\tau\quad \text{ for } f, g : [0, \infty) \to \mathbb{R}$$

As you can see the two definitions are actually equivalent under that particular condition.

The main point is the support being only the non negative reals.

This occurrence is usual while solving ODE's for $u(t),t>0$ with initial data $u(0)$, as the time is usually though at being a positive quantity.


There are two Laplace transforms. One is the one sided one (the usual one) which is valid only for real valued functions over positive reals i.e. $f:\mathbb{R}^+\to\mathbb{R}$, and is given by $$\mathscr{L}[f(t)]=\int_0^\infty f(t)e^{-st}dt$$ Then we have the two sided version where the integral is over $\mathbb{R}$ and the functions are over $\mathbb{R}$.

Now if $f$ and $g$ are defined over positive reals, do you see that the two notions of convolution agree. Also when applying the Laplace transform for $\cos(t)$, the one sided version is used to get that expression $\frac{1}{s^2+1}$. This means that you are assuming the function is zero for $t<0$.

So long story short, the first definition is the more general one and the second one is specific to what I said above.


If your functions are only non-zero on $[0, \infty)$ (in other words zero for all negative values), then the second definition

$$(f \ast g)(t) := \int_0^t f(t - \alpha) g(\alpha) \,d\alpha$$

is equivalent to the first. $g(\alpha)$ will be $0$ for any $\alpha < 0$, so we can drop the $\int_{-\infty}^0$ part. Similarly, $f(t-\alpha)$ will be $0$ for any $\alpha > t$, so we can drop the $\int_t^\infty$ part.


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