integral of delta function of x^2 The name says what I need to calculate. When trying to integrate I stumble upon interpretation problem
$$
\int\limits_{-\infty}^{+\infty} \delta(x^2) dx = \{y=x^2\} = 2\int\limits_{0}^{+\infty} \delta(x^2) \frac{dx}{dy}dy = \int\limits_{0}^{+\infty} \delta(y)\frac{dy}{\sqrt{y}}
$$
My questions are: 
1) How should I interpret the result when zero of delta falls on the limit of integration?
2) If I to ignore reasoning of (1) the answer seems to be $\infty$. Is this true?
 A: The problem here is a problem of definition. 
First of all, by definition the dirac delta function is (as a distribution with finite support) a linear form defined on $\mathcal{C}^0(\mathbb{R})$ that to each function $\phi$ associates $$\int_{-\infty}^\infty\delta(x)\phi(x)dx:=\phi(0).$$ 
Please note that there is no definition of $\delta(x^2)$ other than by the means of change of variable !
To test what $\delta(x^2)$ should be you need to apply everything to a test function :
$$\int_{-\infty}^{+\infty}\delta(x^2)\phi(x)dx=\int_{-\infty}^\infty\delta(y) \frac{\phi(\sqrt{y})}{\sqrt{y}} 1_{y\geq 0} dy$$
BUT the function you wish to test $\delta$ against is not in general continuous at the point $0$, even for $\phi$ smooth with compact support so this makes no sense. 
A: Another view on this integral is: 
$$
\int\limits_{-\infty}^{+\infty} \delta(x^2 - a^2) f(x) dx = 2\int\limits_{0}^{+\infty} \frac{1}{2a} \delta(x) f(x)dx = \frac{f(0)}{a},
$$
which behaves as $1/a$ as $a \rightarrow 0$.
A: $\delta(f(x))$ is only defined if $\nabla f \neq 0$ wherever $f(x) = 0$. In this case, yes it would be "infinite", which is why it is not a well defined object.
Generally we assign the interpretation that
$$\delta(f(x)) \equiv \sum_{f(x_i)=0} \frac{1}{|\nabla f(x_i)|}\delta(x-x_i)$$
