How to compute the integral of two Dirac delta function and another function? How to compute this integral?
$$
\int_{-\infty}^{\infty} \delta (x-a) \delta (x - b) \phi(x)dx
$$
which can be a PDF of a distribution.
I found that
$$
\begin{aligned}
\int_{-\infty}^{\infty}\delta^{n}(x - a)f(x)dx = (-1)^{n} \{ {\frac{d^n}{d^{n}x}}  f(x)|_{x=a}\}
\end{aligned}
$$
from answer from @Lelouch.D.Light
and 
$$
\begin{aligned}
\int_{-\infty}^{\infty}\delta(x - a)f(x)dx = f(a)
\end{aligned}
$$
from Wolfram
How to combine these tools to compute the integral of the expression in the question? Thanks.
 A: If $a$ and $b$ are constants, I'm not aware of any reasonable way to even make sense of $\delta(x-a) \delta(x-b)$, whether by itself or in the context of such an integral.
If you really obtained such a thing, then you've done something wrong.
If $a$ is a variable, one could make sense of such a thing if you say you're computing an integral over distributions in $a$: the result is
$$ \int_{-\infty}^{\infty} \delta (x-a) \delta (x - b) \phi(x) \mathrm{d}x 
= \delta(a-b) \phi(b) $$
The same is true if we computed it as a distribution in $b$ instead. Note that either way, $\delta(a-b) \phi(b) = \delta(a-b) \phi(a)$.
The reason we say this is because of the result we get when integrating these distributions against a test function $g(a)$:
$$ \begin{align}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \delta (x-a) \delta (x - b) \phi(x)g(a) \, \mathrm{d}a \mathrm{d}x 
&=  \int_{-\infty}^{\infty} \delta (x - b) \phi(x)g(x)  \mathrm{d}x 
\\&=  \phi(b)g(b)
\end{align}$$
$$ \int_{-\infty}^{\infty}\delta(a-b) \phi(b) g(a)  \, \mathrm{d}a = \phi(b) g(b)$$
Since both distributions give the same result when integrated against a test function, they are the same distribution.
Pay attention to the order of the integrals in the first line. There are lots of issues with generalized functions, and the very notation we use is one of them! The notation we use doesn't agree with the properties that distributions have, so we have this strange result that the order of integrals get reversed. 
Finally, let me emphasize that this end result is only for the case where $a$ or $b$ (or both!) are variables to be integrated over. This formula is nonsense if you plug in a number for both $a$ and $b$.
