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How to proof the latter equality of, $$ f(a) = \int f(x)\,\delta(x-a)\,dx =\int f(x)\,\delta(a-x)\,dx. $$

kfs
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    It depends on how you define $\delta(f(x))$. There is a canonical definition of $\delta(x)$, and there is a canonical result for $\delta(f(x))$, but you must first define $\delta(f(x))$ before you can derive this result. Some people for instance might adopt the definition $\delta(f(x))=\sum_i \frac{\delta(r_i)}{|f'(r_i)|}$ where $r_i$ are the roots of $f$ which are assumed to be simple. In this case the result you want is trivial. But with another definition it would be nontrivial. – Ian Apr 27 '16 at 19:38
  • The way for the green's function method for solving ODEs? – kfs Apr 27 '16 at 19:47
  • As a Heuristic: The Dirac delta function has meaning only near where it's argument is zero. So changing the absolute value of the argument should not (normally) make a difference. But there are weird ways to define it. – amcalde Apr 27 '16 at 19:48
  • Most likely what you are being expected to do is "pretend that $u$ substitution makes sense", so in $\int f(x) \delta(x-a) dx$ substitute $u=x-a$ to get $\int f(a+u) \delta(u) du=f(a)$ and proceed analogously for $\int f(x) \delta(a-x) dx$ (in this case you get a minus sign which you cancel out by reversing the limits of integration). But this calculation is really not a proof because you haven't actually defined $\delta(f(x))$. – Ian Apr 27 '16 at 19:50
  • Typo above: I meant to say $\delta(f(x))=\sum_i \frac{\delta_{r_i}(x)}{|f'(r_i)|}$, where $\int_{-\infty}^\infty \delta_a(x) g(x) dx = g(a)$. – Ian Apr 28 '16 at 02:28

3 Answers3

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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 $1$.

Jean Marie
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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 $ .

user45664
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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.

Mark Viola
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