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It is not clear to me how this derivation proceeds through the steps. Could someone help me understand how to arrive at this result or point me towards a resource that explains these algebraic manipulations within the Dirac Delta function?

Thanks in advance

Edit: Apologies, not sure how I forgot to mention that P is the primitive function/antiderivative of p.

For context this is related to the Sudakov veto algorithm used in generating parton showers.

1 Answers1

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This is pretty grotesque.

You need to first know a property of the Dirac delta distribution. It is stated as $$\tag1\delta[g(x)]=\frac{\delta(x-x_0)}{|g^\prime(x_0)|}$$ on Wikipedia, say, for any function $g$ that has a simple isolated root at $x_0$, and we will be using this property in reverse for many times.


There are many skipped steps, even at the front. $$ \begin{align} \tag21&=\int_0^1\mathrm d\rho\\ \tag3 &=\int_0^1\mathrm d\rho\int\mathrm dt\ \delta(t-c) \end {align} $$ where the integral of $t$ just needs to include $c$; it is then sensible to choose any $c$ we like, which is in particular $c=P^{-1}[P(u)+\ln\rho]$ that the text used.

We can now apply the property above backwards twice, once on the CDF function $P$ and another time on the exponential $\mathrm e$, leading to the 2nd to last line. By the FTC, $P(t)-P(u)=-\int_t^u p(x)\mathrm dx$

The last line is then a swapping of the integrals, since we are free to use the Dirac delta distribution to destroy $\rho$ rather than $t$, and to do that properly, you also have to consider the integration region.