I found the following transformations on wikipedia's article about variance swap pricing:
My question is: how this is mathematically valid? What is the proper mathematical reasoning behind such transformations? What does $\frac{dS_t}{S_t}$ really mean and how can we "take integral" of it? How can all of this be written rigourously?
We have $$ \frac{dS_t}{S_t} = \mu dt + \sigma dZ_t $$ we have a mathematical tool at our disposal, Ito's formula or Calculus to solve such equations.
The main aim for using such a transform is to get simple integrals on the r.h.s (A physicist approach)
$$ df(S_t) = dS_t\frac{\partial f}{\partial S_t} + \frac{dS_t^2}{2}\frac{\partial^2 f}{\partial S_t^2} $$ where $dS_t$ we already have an equation for $$ df = (\mu S_t dt + \sigma S_t dW_t)\frac{\partial f}{\partial S_t} + \frac{\left(\mu S_t dt + \sigma S_t dW_t\right)^2}{2}\frac{\partial^2 f}{\partial S_t^2} $$ we know that we have quadratic terms in $dW_t$ are kept but terms higher than $dt$ are not. We should also note that $dt = (dW_t)^2$ thus we have $$ df = (\mu S_t dt + \sigma S_t dW_t)\frac{\partial f}{\partial S_t} + \frac{\sigma^2 S_t^2}{2}dt\frac{\partial^2 f}{\partial S_t^2} $$ collecting terms $$ df = \left(\mu S_t \frac{\partial f}{\partial S_t} + \frac{\sigma^2 S_t^2}{2}\frac{\partial^2 f}{\partial S_t^2}\right)dt + \sigma S_t dW_t \frac{\partial f}{\partial S_t} $$
We can trial a solution for $f(S_t) = \log (S_t)$ $$ d(\log (S_t)) = \left(\mu S_t \cdot\frac{1}{S_t} + \frac{\sigma^2 S_t^2}{2}\cdot\frac{-1}{S_t^2}\right)dt + \sigma dW_t S_t \cdot \frac{1}{S_t} $$ or $$ d(\log (S_t)) = \left(\mu -\frac{\sigma^2}{2}\right)dt + \sigma dW_t $$ Which is a nice total integral. We can replace some of the terms on the r.h.s with the original sde. $$ d(\log (S_t)) = \frac{dS_t}{S_t} -\frac{\sigma^2}{2}dt $$ Now lets take the time weighted integral $$ \frac{1}{T}\int_0^Td(\log (S_t)) = \frac{1}{T}\int_0^T \frac{dS_t}{S_t} -\frac{1}{T}\int_0^T \frac{\sigma^2}{2}dt $$ the integral on the left is straight forward $$ \frac{1}{T} \left[\log(S_t)\right]_0^T = \log (S_T) - \log(S_0) = \frac{1}{T}\int_0^T \frac{dS_t}{S_t} - \text{variance} $$ or $$ \text{variance} = -\log\left(\frac{S_T}{S_0}\right)+ \frac{1}{T}\int_0^T \frac{dS_t}{S_t} $$
So in closing assuming that we have Ito at our disposal - he was a mathematician and we know quants are some very bright people. I can safely say that this transforms are ok.
What is strange to me is the fact that $\sigma$ is a constant so why not integrate and leave $\sigma^2$ - the integral is important when we have $\sigma = \sigma(t)$ or something.