How to simplify $\delta(x^2)$ How to simplify the dirac delta of squared x?
How should i solve $\int f(x)\delta(x^2)dx$ for an enough smooth function $f$?
Is this related to $\int f'(x) \delta(x) dx$ ?
I can understand symbols like $\delta(x)$ and $\delta'(x)$ and $\delta(g(x))$ with $g$ having a root $x_0$ with $g'(x_0)\neq 0$ but i can't understand it when $g'(x_0)$ is zero: $\delta(x^2)$,$\delta(|x|)$, etc.
 A: This is a very good question.  Let's generalize the regularization approach that Sangchul presented.  
Let $\delta_n(x)$ be a sequence of (at least piece-wise smooth) positive-valued functions such that
$$\lim_{n\to \infty}\delta_n(x)=\begin{cases}0&,x\ne 0\\\\\infty&,x=0\tag1\end{cases}$$
and for each $n$ 
$$\int_{-\infty}^\infty \delta_n(x)\,dx=1\tag2$$
and for all suitable test functions, $\phi(x)$, we have
$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx=\phi(0)\tag2$$
We say that in distribution $\delta(x) \sim \lim_{n\to \infty}\delta_n(x)$.  This is the Dirac Delta distribution and $\delta_n(x)$ is a regularization thereof.
Now, let's examine whether $\lim_{n\to \infty}\delta_n(x^2)$ has meaning in the distributional sense.  We can write
$$\begin{align}
\int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx&=\int_0^\infty \delta_n(x)\left(\frac{\phi(\sqrt{x}\,)+\phi(-\sqrt{x}\,)}{2\sqrt{x}}\right)\,dx\\\\
&=\int_{-\infty}^\infty  \delta_n(x)\left(\frac{\phi(\sqrt{|x|}\,)+\phi(-\sqrt{|x|}\,)}{2\sqrt{|x|}}\,H(x)\right)\,dx
\end{align}$$
where $H(x)$ is the Heaviside function.  Letting $n\to \infty$, we find for suitable test functions (smooth and of compact support)
$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx=\begin{cases}0&,\phi(0)=0\\\\\infty&,\phi(0)>0\\\\-\infty &,\phi(0)<0\end{cases}$$
So, $\delta(x^2)$ has meaning in distribution on the space of $C^\infty_C$ functions that are $0$ at the origin and the distribution assigns the value of $0$ to each of these functions.  As Sangchul wrote, this is quite a "boring" distribution.

Now, if $\phi(x)$ is a test function such that $\phi(0)=0$, smooth a.e. except at $0$ where  $\lim_{x\to 0^{\pm}}\phi'(x)=C^{\pm}$, then we see that 
$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx=\left(\frac{C^+-C^-}{2}\right)\,H(0)$$
Inasmuch as $H(0)$ is not uniquely defined, then we cannot uniquely define $\delta(x^2)$ as a distribution acting on such functions.  For example, if $\phi(x)=|x|$, then $C^\pm=\pm1$ and $\delta(x^2)\sim H(0)$, which is not uniquely defined.  
A: Let $\delta_n (x) = \frac{n}{2}\mathbf{1}_{[-1/n,1/n]}(x)$. Then $\delta_n$ converges to $\delta$ in distribution sense. Let $f$ be smooth. Then by the Taylor's theorem, we have $f(x) = f(0) + f'(0)x + \mathcal{O}(x^2)$ near $x = 0$. So it follows that
\begin{align*}
\int_{\mathbb{R}} f(x)\delta_n(x^2) \, dx
&= \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} f(x) \, dx \\
&= \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} \left( f(0) + f'(0)x + \mathcal{O}\left(\frac{1}{n}\right) \right) \, dx \\
&= f(0)\sqrt{n} + \mathcal{O}\left(\frac{1}{\sqrt{n}}\right).
\end{align*}
So if $f(0) \neq 0$, then this integral diverges as $n\to\infty$, and if $f(0) = 0$, then this integral converges to $0$ as $n\to\infty$. This tells that $\delta(x^2)$ is either meaningless or boring when paired with nice functions.
On the other hand, if $f(x) = |x|$ then
\begin{align*}
\int_{\mathbb{R}} f(x)\delta_n(x^2) \, dx
= \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} |x| \, dx
= \frac{1}{2}
\end{align*}
This suggests that $\delta(x^2)$ will capture the jump discontinuity of $f'$. Of course, making this heuristics into a meaningful statement would necessitate an appropriate theory of distributions on 'not-so-regular functions', which I do not know much.
