Help understanding copula version of Spearman rank correlation I'm reading through this article (http://www.sciencedirect.com/science/article/pii/S0047259X06000662) where they have a population version of the Spearman rank correlation. I'm having a little bit of trouble understanding the author's derivation, which I reproduce as follows:
"For a two-dimensional random vector $X$ $=(X_{1},X_{2})'$ with distribution function $F$, univariate marginal distributions $F_{X_{1}}$, $F_{X_{2}}$ and a copula $C$ is defined by
$\rho_{S}=\frac{cov(F_{X_{1}}(X_{1}), F_{X_{2}}(X_{2}))}{\sqrt{var(F_{X_{1}}(X_{1}))}\sqrt{var(F_{X_{2}}(X_{2}))}}=$
$= \frac{cov(U_{1},U_{2})}{\sqrt{var(U_{1})}\sqrt{var(U_{2})}}=$
$=\frac{\int_{0}^{1}\int_{0}^{1}uv dC(u,v)-(\frac{1}{2})^{2}}{\sqrt{\frac{1}{12}}\sqrt{\frac{1}{12}}}=$
$=12\int_{0}^{1}\int_{0}^{1}C(u,v)dudv-3$
I understand every step with the exception of the last one. Where did  the 3 come from? And why were they able to change the integrand from $uvdC(u,v)$ to $C(u,v)dudv$? 
I'm trying to teach myself some basic copula theory so there are a few holes in my knowledge here and there.
 A: I came to your question after looking for the same proof and not finding it anywhere. Following StijnDeVuyst suggestion, I managed to prove it using integration by parts. I will omit the integration limits, just keep in mind they exist.
$$\int \int uv \ dC(u, v) = \int\int uv C'(u,v) \ dudv,$$ where $C'(u,v) = \frac{\partial^2}{\partial u \partial v}C(u,v)$.
\begin{equation}
\tag{1}
\label{main}
\int\int uv C'(u,v) \ dudv = \int v \left(\int u C'(u,v) du \right) dv.
\end{equation}
Solving the inner integral
\begin{align*}
\int_0^1 u C'(u,v) du &= u\frac{\partial}{\partial v}C(u,v) \big\rvert_{u=0}^1 - \int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\
&= 1-\int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\
\end{align*}
where the first equality is integration by parts on $u$ and the last equality comes from the fact that $C(1, v) = v$.
Substituting in \ref{main} we have
\begin{align*}
\int \int uvC'(u,v) \ dudv &= \int v \left(1-\int \frac{\partial}{\partial v} C(u,v) \ du \right)dv\\
&= \int_0^1 v \ dv - \int\int v \frac{\partial}{\partial v}C(u,v) \ dudv\\
&= \frac{1}{2}-\int\left(\int v \frac{\partial}{\partial v} C(u,v) \ dv \right) du. \tag{2}\label{segunda}
\end{align*}
Solving the inner integral with integration by parts on $v$ we get
$$ \int_0^1 v \frac{\partial}{\partial v} C(u,v) \ dv = u - \int_0^1C(u,v) \ dv$$
and substituting in \ref{segunda} we have (now including the integration limits)
\begin{align*}
\int_0^1\int_0^1 uv \ dC(u,v) &= \int_0^1 \int_0^1 uvC'(u,v) \ dudv\\
 &= \frac{1}{2} - \int_0^1 u \ du + \int_0^1 \int_0^1 C(u,v) \ dv du\\
&= \int_0^1 \int_0^1 C(u,v) \ du dv
\end{align*}
which proves the identity
\begin{align*}
\rho_S &= 12\int_0^1 \int_0^1 uv \ dC(u,v) - 3\\
&= 12\int_0^1\int_0^1 C(u,v) \ dudv - 3.
\end{align*}
This proof assumes that the copula $C$ has density given by $\frac{\partial^2}{\partial u\partial v}C(u,v)$. When this is not the case, check Theorem 5.1.1 and Corollary 5.1.2 from An Introduction to Copulas by Roger B. Nelsen.
