Proofs of dirac delta property How would I formally prove this property of dirac delta?
$$\int \delta(a-x) \delta(x-b) \,dx = \delta(a-b) $$
I attempted to use the definition of a dirac delta
$$\int f(x)\delta(a-x)\delta(x-b)\,dx=\int f(x)\delta(a-b)\,dx$$
let $g(x)=f(x)\delta(a-x)$
$$\int f(x)\delta(a-x)\delta(x-b)\,dx=\int g(x)\delta(x-b)\,dx$$
let $u=x-b,\:x=u+b, \:du=dx$ 
$$\int g(x)\delta(x-b)\,dx=\int g(u+b)\delta(u)\,du=g(0+b)=g(b)$$
but I couldn't resolve $\frac{d\,g(x)}{dx}$ when I did the u sub
I also couldn't solve the integral of $\int f(x)\delta(a-b)\,dx=\delta(a-b)\int f(x)\, dx$
any hint would be appreciated
 A: Let $\mu = x-b$. Then you have that $$ \begin{align*} 
\int_{-\infty}^{\infty}\delta(a-x)\delta(x-b)dx &= \int_{-\infty-b}^{\infty-b}\delta(a-b-\mu)\delta(\mu)d\mu \\ 
& = \int_{-\infty}^{\infty}\delta(a-b-\mu)\delta(\mu)d\mu \\
& = \delta(b-a) \end{align*} $$
The '=' comes from the contraction of Dirac Functions. This result show that the contraction of two continuous Dirac functions is equivalent to a third Dirac function. 
A: To be completely rigorous, note that in general the product of two measures/distributions/generalised functions is not well-defined; but the convolution is, which is what we have in this case (the convolution of $\delta(x)$ with $\delta(x-b)$ "evaluated" at $a$). Since the result is a measure, the equality just means that the integral against any function is the same, so we check this:
$$\int f(a) \int \delta(a-x) \delta(x-b)\ dx\ da = \int\left(\int f(a)\delta(a-x) da\right) \delta(x-b) dx \\ =\int f(x) \delta(x-b) dx=f(b)=\int f(a)\delta(b-a) da.$$
The interchange of integration here is essentially the definition of the convolution of measures - $\int \delta(a-x) \delta(x-b) dx$ is defined to be the measure such that this manipulation is valid.
A: I presume you know that $$\int f(x)\delta(x-a)\; dx = f(a).\tag{1}$$
Then put $f(x) = \delta(x-b)$ and you get  $$\int \delta(x-b)\delta(x-a)\; dx = \delta(a-b).$$
Maybe this doesn't work because (1) applies only for well-behaved $f$? I forget.  But if not you could make some limit argument based on $\delta(x) = \lim_{a\to\infty} \frac1{\sqrt a}e^{-x^2/a^2}$ and it would go through all right.
