How to change the sign of the Dirac delta function argument? How to proof the latter equality of,
$$
f(a) = \int f(x)\,\delta(x-a)\,dx =\int f(x)\,\delta(a-x)\,dx.
$$
 A: I have read the different (nuanced) answers. Here is another one.
The property that one needs here is the fact that $\delta$ is an (or "can be considered as) an "even" distribution, which is well understandable if $\delta$ is defined (for example) as the limit of even (gaussian) functions $N_{\sigma}(x)=\dfrac{1}{\sigma \sqrt{2\pi}}e^{-\frac12\frac{x^2}{\sigma^2}}$ when their standard-deviation $\sigma \rightarrow 0$, preserving the fact that their integral is $1$.
A: 
In THIS ANSWER and THIS ONE, I provide primers on the Dirac Delta.

Note that the Dirac Delta, $\delta_a$ is a linear functional such that, for each sufficiently smooth test function $f(x)$ having compact support, we have the mapping
$$\langle f,\delta_a\rangle=f(a)$$
Note that this mapping is equivalent to the integral
$$\begin{align}\langle f,\delta_a\rangle&=-\int_{-\infty}^\infty u(x-a)\frac{df(x)}{dx}\,dx \tag 1\\\\&=f(a)\end{align}$$
where $u(x)$ is the unit step function defined by 
$$u(x)=\begin{cases}1&,x\ge 0\\\\0&,x<0\end{cases}$$
If we were to naively integrate by parts the integral in $(1)$, then we obtain the heuristic result 
$$\langle f,\delta_a\rangle =\int_{-\infty}^\infty f(x)\frac{du(x-a)}{dx}\,dx \tag 2$$
Note that if the right-hand side of $(2)$ were interpreted as a Lebesgue or Riemann integral, then the result would be $0$, not $f(a)$.  The reason is that for $x\ne a$, the derivative of the unit step function is zero.  And the value of the integral is not impacted by the values of the integrand at a single point.  However, in terms of Generalized Functions, $(2)$ defines the Dirac Delta as $\delta(x-a)=\frac{du(x-a)}{dx}$.  
Next, we define the Dirac Delta $\delta_a^{-}$  by the integral
$$\begin{align}
\langle f,\delta_a^{-}\rangle&=\int_{-\infty}^\infty u(a-x)\frac{df(x)}{dx}\,dx \tag 3\\\\
&=f(a)
\end{align}$$
Again, naively integrating by parts the integral in $(3)$, and noting that $u(a-x)=1-u(x-a)$, reveals heuristically that 
$$\begin{align}
\langle f,\delta_a^-\rangle&=\int_{-\infty}^\infty f(x)\left(-\frac{du(a-x)}{dx}\right)\,dx\\\\
&=\int_{-\infty}^\infty f(x)\left(\frac{d\left(1-u(a-x)\right)}{dx}\right)\,dx\\\\
&=\int_{-\infty}^\infty f(x)\left(\frac{du(x-a)}{dx}\right)\,dx\\\\
&=\langle f,\delta_a\rangle
\end{align}$$
Therefore, $\delta_a=\delta_a^{-}$ and we can write 
$$\delta(x-a)=\delta(a-x)$$
as was to be shown.
A: Given the $\delta$ distribution property $\delta(y)$ = $\delta(-y)$, then $\delta(x-a)$ = $\delta(a-x)$, and so $\int f(x)\delta(x-a)dx $ = $\int f(x)\delta(a-x)dx $ .
