Convolution with delta function I am merely looking for the result of the convolution of a function and a delta function. 
I know there is some sort of identity but I can't seem to find it. 
$\int_{-\infty}^{\infty} f(u-x)\delta(u-a)du=?$
 A: The delta "function" is the multiplicative identity of the convolution algebra. That is, $$\int f(\tau)\delta(t-\tau)d\tau=\int f(t-\tau)\delta(\tau)d\tau=f(t)$$
This is essentially the definition of $\delta$: the distribution with integral $1$ supported only at $0$.
A: It's called the sifting property:
$$
\int_{-\infty}^\infty f(x)\delta(x-a)\,dx=f(a).
$$
Now, if
$$
f(t)*g(t):=\int_0^t f(t-s)g(s)\,ds,
$$
we want to compute
$$
f(t)*\delta(t-a)=\int_0^t f(t-s)\delta(s-a)\,ds.
$$
With an eye on the sifting property above (which requires that we integrate "across the spike" of the Dirac delta, which occurs at $a$, we consider two cases.


*

*If $t<a$, then $\delta(s-a)\equiv 0$ since $0\le s\le t<a$. Therefore $\int_0^t f(t-s)\delta(s-a)\,ds\equiv 0$.

*If $t\geq a$, then by the sifting property, $\int_0^t f(t-s)\delta(s-a)\,ds=f(t-s)\Big|_{s=a}=f(t-a)$.
Thus,
\begin{align}
f(t)*\delta(t-a)=\int_0^t f(t-s)\delta(s-a)\,ds&=\begin{cases} 0, &t<a,\\ f(t-a) &t\ge a,\end{cases}=f(t-a)u(t-a),
\end{align}
where $u(t)$ denotes the unit step function.
