# Integral of two Dirac delta

Is it possible to assign a meaning to $$\int_{-\infty}^\infty dt \; \delta(a-x(t)) \delta(b-y(t))$$

Is it a distribution? Since the usual definition of the Dirac delta involves a single integral for a single Dirac delta $$\int_{-\infty}^\infty dt \; \delta(t-a) f(t) = f(a)$$

using this to do the integral over one of the delta we are left with the other Dirac delta.

For a single Dirac delta it is straightforward to show by variable substitution that $$\int_{-\infty}^\infty dt \; \delta(a-x(t))= \left ( \frac{dx}{dt} \Bigg |_{x(t)=a} \right )^{-1}$$

(assuming that $x(t)$ is injective such that there is only one $t$ where $x(t)=a$).

By some intuitive argument I want this to be the same as: if $x(t)$ and $y(t)$ are both injective and $x(t_1)=a$ and $y(t_2)=b$ then it is $1/(\text{some derivative...})$ if $t_1=t_2$ and $0$ otherwise - but this is not correct, right?

Thanks!

• If $a-x(t)$ and $b-y(t)$ are not zero for the same $t$ then it should be possible. – md2perpe Aug 21 '17 at 21:09
• If they are not zero for the same $t$? Can you elaborate - is it a distribution? – smörkex Aug 21 '17 at 21:14
• No $\delta(t)^2$ doesn't mean anything. But $\delta \ast \delta(t)$ is well-defined as a distribution as well as $T(x,y)=\int_a^b \delta(x-\gamma(t),y-\gamma(t))dt$ where $\delta(x,y)=\delta(x)\delta(y)$ is the Dirac delta in $\mathbb{R}^2$ – reuns Aug 21 '17 at 21:20
• If $a-x(t)$ and $b-y(t)$ are not zero for the samt $t$ then the mapping $\phi \mapsto \int_{-\infty}^{\infty} \delta(a-x(t)) \, \delta(b-y(t)) \, \phi(t) \, dt$ can be given a meaning as a distribution; it will however be the zero distribution. – md2perpe Aug 21 '17 at 21:27
• Two quite relevant links: math.stackexchange.com/questions/2389100/… math.stackexchange.com/questions/2336866/… – md2perpe Aug 21 '17 at 21:28