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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?

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    $\begingroup$ Please, please break yourself of this horrible habit of connecting things with = signs even when you don't mean that they are equal. $\endgroup$ – Paul Sinclair Oct 30 '19 at 16:36
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$\delta(f(x))$ 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

$$\delta(f(x)) \equiv \sum_{f(x_i)=0} \frac{1}{|\nabla f(x_i)|}\delta(x-x_i)$$

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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.

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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 $a \rightarrow 0$.

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