I am trying to understand a substitution in a paper I am reading. The paper is "The Application of Linear Filter Theory to the Direct Interpretation of Geoelectrical Resistivity Sounding Measurements" by D.P. Ghosh. I am getting caught up on understanding how eqns 4 - 9 are derived. The equations are as follows:
$$ \rho_a(s) = s^2 \int_{0}^\infty T(\lambda) J_1(\lambda s)ds \qquad (4)$$ Where $\rho_a$ is the apparent resistivity and $T(\lambda)$ is a resistivity transform function. Using a Hankel Transform, we get: $$ T(\lambda) = \int_{0}^\infty \rho_a(s) J_1(\lambda s)\left[\frac1s\right]ds \qquad (7)$$ New variables are then defined by $$x = ln(s) \quad and \quad y = ln\left(\frac1\lambda\right) \qquad (8)$$ Substituting (8) in (7), we get $$ T(y) = \int_{-\infty}^\infty \rho_a(x) J_1\left( \frac{1}{e^{y-x}} \right)dx \qquad (9)$$ Please note that any missing equations were for another resistivity equation.
I can't figure out how the substitutions in (8) eliminated the $\frac1s$ in (7) and I'm not sure why the limit changes from 0 to $-\infty$. Any help in figuring this out would be great, I'm kind of stuck.
You have copied formula (4) incorrectly from the paper: $d\lambda$ not $ds$ in (4).
Starting with
$$T(\lambda) = \int_{s=0}^{s=\infty} \rho_a(s) J_1(\lambda s)\left[\frac1s\right]ds \qquad (7)$$
and using
$$x=\ln(s) \quad \Rightarrow \quad e^x = s \quad \mbox{and} \quad dx = \dfrac{1}{s}ds$$ $$\ln(0) \rightarrow -\infty \quad \mbox{and} \quad \ln(\infty) \rightarrow \infty$$
you now have
$$T(\lambda) = \int_{x=-\infty}^{x=\infty} \rho_a(x) J_1(\lambda e^x)dx$$
Then using
$$y = \ln\left(\dfrac{1}{\lambda}\right) \quad \Rightarrow \quad e^y = \dfrac{1}{\lambda}\quad \Rightarrow \quad {\lambda} = \dfrac{1}{e^y}$$
you now have
$$\begin{align} \\ T(y) &= \int_{-\infty}^\infty \rho_a(x) J_1\left(\dfrac{1}{e^y} e^x\right)dx\\ \\ &= \int_{-\infty}^\infty \rho_a(x) J_1\left(\dfrac{1}{e^{y-x}}\right)dx \quad (9)\\ \end{align}$$