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\begin{equation} \int\delta(E^2-\mathbf{p}^2-m^2)dE=\frac{1}{2E_\mathbf{p}} \end{equation}

Shouldn't integrating the delta function like this just give 1?

Qmechanic
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Luka Milic
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    Have a look at e.g. this question. And your integration variable shouldn't be $E$ - how should $E$ be able to occur on the RHS? – ACuriousMind May 25 '15 at 14:23
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    $\delta(f(x)) = \sum_k\frac1{|f'(x^_k)|}\delta(x-x^_k)$ where the $x^*_k$s are the zeroes of $f$...and you are probably integrating over positive energies alone... – Phoenix87 May 25 '15 at 14:23
  • @ACuriousMind Yeah I don't like it either hence why I'm asking if anyone can clarify what its trying to say – Luka Milic May 25 '15 at 14:27
  • Its from some lecture notes, here is the context: http://imgur.com/HPg1GZT – Luka Milic May 25 '15 at 14:45
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    The identity given by @Phoenix87 above can be found in Kusse & Westwig's Mathematical Physics: Applied Mathematics for Scientists and Engineers (among other places, I'm sure, but that's where I know where to find it.) It can also be proven by splitter the original integral up into small regions about each zero $x^*_k$, requiring that $f$ be invertible in each such small region, and then changing variables to integrate over $f$ instead of $x$ in each region. – Michael Seifert May 25 '15 at 15:36

1 Answers1

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With the help from the comments this now makes sense.

\begin{equation} \int \delta(E^2-p^2-m^2)dE \end{equation} With $$E_p^2-p^2-m^2=0$$ Use substitutions \begin{equation} f(E)=E^2-p^2-m^2\quad df=2EdE \end{equation} \begin{equation} \int \delta(f)\frac{df}{2E(f)}=\frac{1}{2E_p} \end{equation} $E(f)$ is easily found by inverting $f$

Thanks!

Luka Milic
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