I've come into a bit of a snag, and thought some more talented mathematicians could maybe help. I am trying to do the following integral:

$$S(x,t) = \int I(z)\delta(x-G(z,t)) \mathrm{d}z,$$

where $G(z,t)$ is a function which 'pushes' the original function $I(z)$ into $S(x,t)$ at some later time. I've tried using some Dirac delta identities but have not had much success. Any help would be very much appreciated. Thank you.


Have you tried to use the decomposition of the "composite" delta function $\delta(f(x))$,


In your case you have $$\delta(x - G(z,t)) = \sum_{i} \frac{\delta(z-z_i)}{|\partial_z G(z,t)|_{z=z_i}|}$$ where the sum goes over the solutions $z_i(x,t)$ of the equation $G(z,t) = x$, so that $$S(x,t) = \sum_{i} \frac{I(z_i)}{|\partial_z G(z,t)|_{z=z_i}|}$$ Potential problems might arise at points where $\partial_z G(z,t)|_{z=z_i} = 0$.


You'd think of $\int f(x) \delta(x) \; \mathrm{d} x = f(0)$ as picking just the value of $f$ where $\delta$'s argument is zero. I.e., in this case the result is $I(z)$ wherever $G(z, t) = x$.

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    $\begingroup$ Aren't you missing a factor $\partial G/\partial z$? Also, if there are several values of $z$ for which $G(z,t)=x$, then shouldn't you be adding together all the corresponding values $I(z)$? $\endgroup$ – Andreas Blass Oct 27 '14 at 16:44

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