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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.

$\int_{-\infty}^{\infty} f(u-x)\delta(u-a)du=?$

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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.

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

  2. 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.

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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 $0$.

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