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I am trying to understand an integral I saw in a book. The book is "Earth Resistances" by G.F. Tagg. Unfortunately it's difficult to get a hold of this book, but I am confused by equation 3.9 is in chapter 3 of the book. I will put what I think is necessary in this but if anyone feels as though there is missing information please tell me.

The following equation is for the potential measured at the surface given a model of the earth that has 2 soil layers of different resistivity. The following equation focuses on only the part that I am stuck on.

$$V_1 = \int_0^\infty \left[\frac{2ke^{-2\lambda h}}{1 - ke^{-2\lambda h}}\right] J_0(\lambda r) d\lambda$$ The fraction inside of the brackets is then expanded into it's geometric series: $$\frac{2ke^{-2\lambda h}}{1 - ke^{-2\lambda h}} = 2\sum_{n=1}^{n=\infty} k^n e^{-2n\lambda h}$$ This yields $$\int_0^\infty \sum_{n=1}^{n=\infty} k^n e^{-2n\lambda h} J_0(\lambda r) d\lambda = 2\sum_{n=1}^{n=\infty} \frac{k^n}{\sqrt{r^2 + (2nk)^2}}$$ It's the last equivalence that is confusing me. I cant find a Laplace transform that has the same form as the RHS. The one I can find is this, which as far as I can tell doesn't reduce into the RHS equation. Am I missing something?

R-dubs
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That should be $\sqrt{r^2 + (2nh)^2}$ on the right side, not $\sqrt{r^2 + (2nk)^2}$.

$$ \int_0^\infty \exp(-p \lambda) J_0(r \lambda)\; d\lambda = \dfrac{1}{\sqrt{p^2+r^2}}$$ if $p, r > 0$. Take $p = 2 n h$.

Robert Israel
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