Integrating the dirac delta function with a periodic argument

Question

For example $$\int_0^\infty \frac{\delta \big (\cos(x^2) \big)}{x^2} dx$$

I'm just not sure how to handle there being multiple solutions for $\delta (cos(x^2))=0$

Answer 1

EDIT: Because there have been wrong statements posted, I add a general instruction. For a differentiable function $g(x)$ with zeros $x_n$ of order 1, one defines $$\delta (g(x)) = \sum_{n} \frac{1}{\vert g'(x_n) \vert} \delta (x - x_n)$$ Check out e.g. Wikipedia. The reason why one chooses this definition is because it is practical. Check out the two approaches below that give the same result to get an idea why this definition is 'right'.

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First of all, I cast $\delta (\cos x^2 )$ into standard form: The zeros of $\cos x^2$ are $x^2 = \pi (n+\frac 1 2 ), n=0, 1, \dots$. I.e. for every $n$, there is precisely one positive zero. The derivatives are: $$\frac{d}{dx} \cos x^2 = 2 x \sin x^2$$ For $x = \sqrt{\pi (n+\frac 1 2)}$, this gives $$2 \sqrt{\pi (n+\frac 1 2)} \cdot (-1)^{n+1}$$ Thus, $$\int_0^\infty dx\,\frac{\delta (\cos x^2)}{x^2} = \int_0^\infty dx\,\sum_{n=0}^\infty \frac{1}{2 \sqrt{\pi (n+\frac 1 2 )}} \delta (x- \sqrt{\pi (n+\frac 1 2)}) \frac{1}{x^2} =\\= \sum_{n=0}^\infty \int_0^\infty dx\, \frac{1}{2 \sqrt{\pi (n+\frac 1 2 )}} \delta (x- \sqrt{\pi (n+\frac 1 2)})\frac{1}{x^2} = \sum_{n=0}^\infty \frac{1}{2 (\pi (n+\frac 1 2 ))^{3/2}}$$

This converges absolutely, as can be seen by evaluating the integral and exploiting monotonicity.

Another way to get the result is by direct substitution: $d(x^2) = 2 x\,dx, \frac{dx}{x^2}=\frac{2 x dx}{2 (x^2)^{3/2}}$

$$\int_0^\infty dx\,\frac{\delta (\cos x^2 )}{x^2} = \int_0^\infty d\xi\,\frac{\delta (\cos \xi )}{2 \xi^{3/2}} $$ Here, one has to account for the zeros $$\xi = \pi (n + \frac 1 2)$$ and thus $$\dots = \frac{1}{2} \sum_{n=0}^\infty \frac{1}{(\pi (n+\frac 1 2))^{3/2}}$$