How do I show the equivalence of the two forms of the Anderson-Darling test statistic? It's stated in many places regarding the Anderson-Darling test statistic, which is defined as
$$n\int_{-\infty}^\infty \frac{(F_n(x) - F(x))^2}{F(x)(1 - F(x))}dF(x)$$
that this is functionally equivalent to the statistic
$$A^2 = -n - S$$
where
$$S = \sum_{k=1}^n\frac{2k-1}{n}\left(\ln F(Y_k) + \ln(1 - F(Y_{n+1-k}))\right)$$
Note that $F_n(x)$ is the empirical distribution function and $F(x)$ is the distribution to which we are comparing the sample. $Y_k$ is the $k^{th}$ ranked element in the sample.
I even went so far as to read the original 1954 paper by Anderson and Darling and I have yet to discover how this equivalence was computed - these authors merely stated the equivalence too. I've tried writing out the numerator inside the integral and splitting into 3 integrals - I was only able to simplify one of them.
I have an inkling that maybe the Probability Integral Transformation should be applied, but I'm not really sure how. I'd really appreciate if anyone could give any pointers.

I think you can prove it by dividing the integral into (n+1) integrals on the intervals $[Y_k; Y_{k+1})$. 
