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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 $\sqrt{2m(E_{i}+\hbar\omega)}$.

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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 $$p_f^2 \left(\frac{d}{dp_f}(p_f^2/2m - E_i - \hbar \omega) \right)^{-1} = p_f^2 (m/p_f) = m p_f.$$

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