approximating an integral/hypergeometric function 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?
 A: 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}
