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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.

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

http://en.wikipedia.org/wiki/Dirac_delta_function#Composition_with_a_function

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 $\partial_z G(z,t)|_{z=z_i} = 0$.

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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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