Integrating a delta function

Question

I'm trying to express a flux $F$ in terms of frequency $\nu$ using:

$$F=\frac{2 \pi}{D^2}\frac{a_0}{\sqrt{a_1\nu}}B^\frac{3}{2}Q\int p^2 dp \int r^2 (\frac{p}{mc})^2p^{-4}dr\;\delta\left[\frac{p}{mc}-\sqrt{\frac{\nu}{B}}\right] $$

Where $p$ is momentum, $r$ is distance from the centre of a sphere, $B$ is the magnetic field, $m$ is mass and $c$ is light speed, $a_0$ is a constant with units W Hz$^{-1}$ T $^{-2}$ and $a_1$ a constant with units Hz T$^{-1}$.

Dealing with the $p$'s first

$$F=\frac{2 \pi}{D^2}\frac{a_0}{\sqrt{a_1\nu}}B^\frac{3}{2}Q\int dp \int r^2 (\frac{1}{mc})^2dr\;\delta\left[\frac{p}{mc}-\sqrt{\frac{\nu}{B}}\right] $$

The integral over $r$, which contains a Dirac delta function:

$$\int r^2 (\frac{1}{mc})^2dr\;\delta\left[\frac{p}{mc}-\sqrt{\frac{\nu}{B}}\right] $$

Leads to

$$F=\frac{2 \pi}{D^2}\frac{a_0}{\sqrt{a_1\nu}}B^\frac{3}{2}Q\frac{r(s)^3-r(a)^3}{3} \int (\frac{1}{mc})^2dp \;\delta\left[\frac{p}{mc}-\sqrt{\frac{\nu}{B}}\right] $$

Where we've integrated $r$ between $s$ and $a$.

The other integral is

$$\int (\frac{1}{mc})^2dp \;\delta\left[\frac{p}{mc}-\sqrt{\frac{\nu}{B}}\right]$$

and is between some initial momentum $p_0$ and a maximum momentum $p_{max}$. How would I deal with the integral over $p$ whilst also expressing the entire equation in terms of $\nu$?

Answer 1

Irrespective of dimensionality problem by changing the variable to $x=\frac{p}{mc}$ one gets: $$F=\frac{r^3}{3}\int p^2 dp\;\delta\left[\frac{p}{mc}-\sqrt{\frac{\nu}{B}}\right]=\frac{r^3m^3 c^3}{3}\int x^2 dx\;\delta\left[x-\sqrt{\frac{\nu}{B}}\right]=\frac{r^3m^3 c^3}{3} \left|\frac{\nu}{B}\right|.$$ Which is being the consequence of Dirac delta translation property: $$\int _{-\infty }^{\infty }f(x)\delta (x-X)\,dt=f(X).$$

Answer 2

First, you make a change of variables: $$u = \frac{p}{mc}-\sqrt{\frac{\nu}{B}},\qquad dp = mc\,du\ .$$ With this, we have \begin{align} \frac{r^3}{3}\int p^2 dp\,\delta\left[\frac{p}{mc}-\sqrt{\frac{\nu}{B}}\right] ={}&\frac{m^3c^3r^3}{3}\int \left(u+\sqrt\frac{\nu}{B}\right)^2 du\,\delta\left[u\right]\\ ={}&\frac{m^3c^3r^3}{3}\frac{\nu}{B}\ . \end{align} That being said, as a remark from a mathematician, one should be aware that the Dirac delta is in fact a distribution, and one should be careful when treating it as if it were a function.