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In the PBRT book section on null scattering, they define the null-scattering coefficient $\sigma_n$ and the constant majorant $\sigma_{maj}$, and define the equation:

$\frac{dL_o(p, \omega)}{dt} = -(\sigma_{maj} - \sigma_n(p, \omega))L_i(p, -\omega) $.

Then, they write that they integrate this equation over the segment of a ray and divide it by $L(p)$ to get:

$T_r(p \to p') = e^{-\sigma_{maj}t} + \int_0^d e^{-\sigma_{maj}t} \sigma_n(p + t \omega) \, T_r(p + t\omega \to p') \, dt$,

I have been trying to derive it myself using the same method but I haven't had much success. This is as close as I was able to get, but I'm not sure if I'm on the right path:

$\int_0^d \, dL(t) = - \int_0^d (\sigma_{maj} - \sigma_n(t)) \, L(t) \, dt$,

$\frac{L(d)}{L(0)} = 1 - \int_0^d (\sigma_{maj} - \sigma_n(t)) \, \frac{L(t)}{L(0)} \, dt$,

$T_r(p \to p') = 1 - \int_0^d (\sigma_{maj} - \sigma_n(t)) \, T_r(p \to p + t\omega) \, dt$,

$T_r(p \to p') = 1 - (\int_0^d \sigma_{maj} \, T_r(p \to p + t\omega) \, dt - \int_0^d \sigma_n(t) \, T_r(p \to p + t\omega) \, dt)$.

Focusing on the first occurrence of the transmittance function under the integral:

$T_r(p \to p + t\omega) = e^{-\int_0^t \, (\sigma_{maj} - \sigma_n(s)) \, ds} $,

$T_r(p \to p + t\omega) = e^{\int_0^t \, \sigma_n(s) \, ds} e^{-\sigma_{maj}t} $.

Substituting into the previous equation:

$T_r(p \to p') = 1 - (\int_0^d \sigma_{maj} \, e^{\int_0^t \, \sigma_n(s) \, ds} e^{-\sigma_{maj}t} \, dt - \int_0^d \sigma_n(t) \, T_r(p \to p + t\omega) \, dt)$

This is where I get stuck, I can't see a way to simplify things to get the equation given in the book. Am I heading in the right direction or is there some trick to getting the correct solution that I'm missing?

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I stumbled upon a derivation that works, although doesn't involve the steps mentioned in the book:

$\frac{dL(p + t\omega)}{dt} = -(\sigma_{maj} - \sigma_n(p + t\omega))L(p + t\omega)$.

$T_r(p + t\omega \to p') = \frac{L(p')}{L(p + t\omega)}$,

$\frac{d}{dt}(T_r(p + t\omega \to p')) = L(p')\frac{d}{dt}(L(p + t\omega)^{-1})$,

$\frac{d}{dt}(T_r(p + t\omega \to p')) = -L(p')L(p + t\omega)^{-2}\frac{d}{dt}(L(p + t\omega))$,

$\frac{d}{dt}(T_r(p + t\omega \to p')) = -\frac{L(p')}{L(p + t\omega)^{2}}\frac{dL(p + t\omega)}{dt}$,

$\frac{d}{dt}(T_r(p + t\omega \to p')) = -\frac{L(p')}{L(p + t\omega)^{2}}(-(\sigma_{maj} - \sigma_n(p + t\omega))L(p + t\omega))$,

$\frac{d}{dt}(T_r(p + t\omega \to p')) = \frac{L(p')}{L(p + t\omega)}(\sigma_{maj} - \sigma_n(p + t\omega))$,

$\frac{d}{dt}(T_r(p + t\omega \to p')) = T_r(p + t\omega \to p')(\sigma_{maj} - \sigma_n(p + t\omega))$,

$\frac{d}{dt}(T_r(p + t\omega \to p')) - \sigma_{maj}T_r(p + t\omega \to p') = - \sigma_n(p + t\omega)T_r(p + t\omega \to p')$.

Multiply by $e^{-\sigma_{maj}t}$:

$e^{-\sigma_{maj}t}\frac{d}{dt}(T_r(p + t\omega \to p')) - \sigma_{maj}e^{-\sigma_{maj}t}T_r(p + t\omega \to p') = - e^{-\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p')$

Applying the product rule:

$\frac{d}{dt}(e^{-\sigma_{maj}t}T_r(p + t\omega \to p')) = - e^{-\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p')$,

$\int_0^d d(e^{-\sigma_{maj}t}T_r(p + t\omega \to p')) = - \int_0^d e^{-\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$,

$e^{-\sigma_{maj}d} - T_r(p \to p') = - \int_0^d e^{ -\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$,

$T_r(p \to p') = e^{-\sigma_{maj}d} + \int_0^d e^{ -\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$.

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