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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.

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2 Answers 2

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 $0$:

$$\delta(g)=g(0)$$

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

$$\phi*g(t)=\phi(g(t-\cdot))$$

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

$$f*g(t)=g(t-c)$$

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

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