I am looking to approximate the following integral for small $z$:
$\int_0^{\infty}dy \frac{1}{z} e^{-y/z} \frac{w e^{-y}}{s + w e^{-y}}$ .
The integral can be solved in general to be a hypergeometric function:
$ \frac{w}{s (z + 1)}\times_2F_1(1, 1 + 1/z; 2 + 1/z; -w/s)$
so expanding that in small $z$ is also an idea (though Mathematica refuses to do so). By numerical trial and error, I have found that the expression
$\frac{w (1-z)}{s + w (1-z)}$
is a decent approximation and I have been trying to prove that to be the case. My attempts have been along the lines that when $z$ is small, $e^{-y/z}$ should decay very rapidly. In the limit of $z$ being zero, I can use a Delta function and get a result that I expect. However adding in a width of $z$ puts me at an impasse.
Any suggestions on how to tackle this problem?
If assuming that the result obtaining from the integral is correct, your question is how to approximate the given hypergeometric function with condition that $z$ is very small.
Applying the Euler's hypergeometric transformation (Mathworld - Wolframalpha, equation (32)) and series expansion (equation (7)) gives \begin{align} S &=\frac{w}{s(z+1)}\,{}_2F_1\left(1, 1+\frac{1}{z}; 2+\frac{1}{z};-\frac{w}{s} \right) \\ &=\frac{w}{s(z+1)}\left(1 + \frac{w}{s}\right)^{-1}\,{}_2F_1\left(1, 1; 2+\frac{1}{z}; \frac{w}{w+s}\right) \qquad (\text{Euler's transformation})\\ &= \frac{w}{(z+1)(w+s)}\, \left[1 + \frac{1\times 1}{1!(2 + 1/z)} \frac{w}{w+s} + \frac{1(1+1)1(1+1)}{2!(2+1/z)(2+1/z+1)}\left(\frac{w}{w+s}\right)^2 + \cdots\right]\\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad(\text{series expansion}) \end{align} Hence, if $z$ is small and $|w/(w+s)| < 1$, all of terms in the series expansion are very small, except the lowest order term. Hence, we can approximate the expression to \begin{equation} S \approx \frac{w}{(z+1)(w+s)}. \end{equation}