Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $c$ be a positive constant and let $f(t)=\delta(t-c)$. Compute $f*g$. So setting up the integral, I get $$ (f*g) = \int_0^t \delta (t-\tau-c) g(\tau) d\tau$$ I am unsure of how to take the integral of the delta function. Any help would be greatly appreciated.

share|cite|improve this question

The delta "function" is strictly speaking not a function, but a distribution, that is a continuous linear functional on a space of test functions. It sends a test function $g$ onto its value at $0$:


You may think of $\delta(g)$ as "$\int \delta(\tau)g(\tau)d\tau$". Be however aware that there does not exist an actual function $\delta$ such that $\int \delta(\tau)g(\tau)d\tau=g(0)$ for every test function $g$.

Now if $\phi$ is any distribution and $g$ is a test function, then


This is motivated by the case when $\phi$ is a function: in that case,

$$\phi(g(t-\cdot))=\int g(t-\tau) \phi(\tau)d\tau$$

Your distribution is the $\delta$ distribution shifted by $c$, that is $f(g)=g(c)$. Therefore


share|cite|improve this answer

If you want compute $f*g$, that is $$(f*g)(t) = \int_{-\infty}^{+\infty}\delta(t-c-\tau)g(\tau)d\tau=g(t-c)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.