# 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?

• Please, please break yourself of this horrible habit of connecting things with = signs even when you don't mean that they are equal. – Paul Sinclair Oct 30 '19 at 16:36

$$\delta(f(x))$$ 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

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

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.

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 $$a \rightarrow 0$$.