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I'm trying to invert:

$f(x)=\frac{E}{4\pi D \mid x \mid}e^{\frac{-\mid x \mid}{\sqrt{DT}}}$

Where E,D and T are just some arbitrary real parameters.

Mathematica ends up with an expression in terms of ProductLog, which is the Lambert W-Function: the inverse of $g(W)=We^W$, that looks like $$\sqrt{DT} ProductLog(\frac{E}{4\sqrt{T} D^{\frac{3}{2}} \pi y})$$

but I'd like to arrive at it myself to see how it does it.

I've only made the obvious steps:

$$y=\frac{E}{4\pi D \mid x \mid}e^{\frac{-\mid x \mid}{\sqrt{DT}}}\Rightarrow\sqrt{\Delta T} log(\frac{rate}{4\pi D y})=log(\mid x \mid)\mid x \mid$$

and I imagine the Lambert W-Function will come in somewhere here.

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Write your equation as $t = e^{-s}/s$, where $s = |x|/\sqrt{DT}$ and $t = 4 \pi D^{3/2} T^{1/2} y$, and then as $s e^s = 1/t$. Thus $s = W(1/t)$.

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  • $\begingroup$ Ah, very neat. Thanks! $\endgroup$ – Mike Dec 22 '16 at 22:18
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Notice that

$$a\log a=(\log a)e^{\log a}$$

where $a=|x|$. Now take the Lambert W function.

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