Integral with $\delta$ function

Question

I got this integral in a quantum field theory problem:

$$ \int\limits_{-\infty}^{+\infty}\!\!\! dp \, \frac{p^2 \delta\left(\sqrt{p^2-m_2^2}+\sqrt{p^2-m_3^2} -m_1\right)}{\sqrt{p^2-m_2^2} \sqrt{p^2-m_2^3}} $$

Are there any 'tricks' to compute integrals where there $\delta$ is composed with another function?

EDIT: The solution is $\frac{\sqrt{(m_1-m_2-m_3) (m_1+m_2-m_3) (m_1-m_2+m_3) (m_1+m_2+m_3)}}{m_1^2}$. I calculated it as proposed in the answers.

Answer 1

Assume that:

• $f$ has at most countably many roots $r_n$
• $f$ is continuously differentiable at $r_n$ for each $n$
• $f'(r_n) \neq 0$ for each $n$.
• $g$ is continuous, at least at $r_n$.

Then

$$\int_{-\infty}^\infty g(x) \delta(f(x)) dx=\sum_n \frac{g(r_n)}{|f'(r_n)|}$$

One can heuristically prove this using a change of variable, but as a strict mathematical matter this should be considered to be a definition.

In your case, $f(p)=\sqrt{p^2-m_2^2}+\sqrt{p^2-m_3^2}-m_1$. This function is increasing for $p \ge \max \{ |m_2|,|m_3| \}$, decreasing for $p \le -\max \{ |m_2|,|m_3| \}$, and not real-valued elsewhere. It is also even. So for fixed $m_2,m_3$ and sufficiently large $m_1$, $f$ has two roots, both of which satisfy the assumptions above.

Answer 2

We have (see here) $$ \delta\left(\sqrt{p^2-m_2^2}+\sqrt{p^2-m_3^2} -m_1\right) = \delta(g(p)) = \sum_k \frac{\delta(p-p_k)}{\lvert g'(p_k)\rvert} $$ where $$ g(p) = \sqrt{p^2-m_2^2}+\sqrt{p^2-m_3^2} -m_1 $$ which has simple roots at $p_k$: $$ g(p_k) = 0 $$