# Question on doing the integral for Fermi golden rule

Today in the lecture, my professor did something which confused me

As an example, we consider the photoelectric effect, in which an electron bound in a Coulomb potential is ionized after interacting with an external electromagnetic field. [...] The absorption rate is given by $$\Gamma_{i\rightarrow f}=\frac{4a_{0}^{3}e^{2}}{m^{2}\pi\hbar^{4}c^{2}}\frac{(\mathbf{A}_{0}\cdot\mathbf{p}_{f})^{2}}{(1+(p_{f}a_{0}/\hbar^{2})^{4}}\delta(p_{f}^{2}/2m-E_{i}-\hbar\omega)$$

This gives us the rate for a precise final momentum $p_{f}$ . Typically, what we want to know is the rate of electrons detected in a solid angle $\mathrm{d}\Omega$ $$\frac{\mathrm{d}\Gamma}{\mathrm{d}\Omega}=\int_{0}^{\infty}p_{f}^{2}\,\mathrm{d}p_{f}\,\Gamma_{i\rightarrow p_{f}}$$

Doing the $p_{f}$ integral, we have $$\frac{\mathrm{d}\Gamma}{\mathrm{d}\Omega}=\frac{4a_{0}^{3}e^{2}p_{f}}{m\pi\hbar^{4}c^{2}}\frac{(\mathbf{A}_{0}\cdot\mathbf{p}_{f})^{2}}{(1+(p_{f}a_{0}/\hbar^{2})^{4}}$$

where $p_{f}=\sqrt{2m(E_{i}+\hbar\omega)}$.

but I don't get how the integral just multiplies a term $m \, p_f$ to the first expression. I thought that the $\delta$ function replace $p_f$ with $\sqrt{2m(E_{i}+\hbar\omega)}$.

The delta function $\delta(x)$ has unit area, but the function $\delta(2x)$ is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is $$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$ where the $x_i$ are the roots of $f$. In this case the factor you pick up is $$p_f^2 \left(\frac{d}{dp_f}(p_f^2/2m - E_i - \hbar \omega) \right)^{-1} = p_f^2 (m/p_f) = m p_f.$$