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

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. 
