I am having trouble evaluating the following integral.
Here, $F_{X}(x)$ and $f_{X}(x)$ are the CDF and PDF of an arbitrary continuous random variable $x$ (with domain $(0,1)$). (And so I am trying to calculate the expectation of the order statistic $(L)$ of $F_{X}(x)$.
\begin{equation} E[f_{x_{(L)}}]=\int_{0}^1 \frac{(2L-1)!}{(L-1)!(L-1)!}(F_{X}(x))^{L}(1-F_{X}(x))^{L}f_{X}(x) x dx \end{equation}
\begin{equation} E[f_{x_{(L)}}]= \frac{(2L-1)!}{(L-1)!^2} \int_{0}^1(F_{X}(x))^{L}(1-F_{X}(x))^{L}f_{X}(x) x dx \end{equation}
Call $\frac{(2L-1)!}{(L-1)!^2}=J$, and the mean of $F_{X}(x)$ $\mu$. Setting $v=f_{X}(x) x$and $u=(1-F_{X}(x))^{L}F_{X}^{L}$, I can integrate by parts: \begin{eqnarray} J\int uv dx = J[u\int v dx -\int u' (\int v dx) dx]\\ =J[u \mu - \int u' \mu dx] \\ =J[u \mu -\mu \int u' dx] \\ =J[u\mu - \mu u] \\ =0 \end{eqnarray} This is clearly non-sensical and so I assume I have made a mistake but I am not sure where!
You replaced the second occurrence of $\int v\,\mathrm dx$ by $\mu$, but $\mu$ is the definite integral from $0$ to $1$, whereas this is the indefinite integral. Your confusion may be related to the invalid notation, in which you denoted both the outer and the inner integration variable by $x$. Things should readily be cleared up if you distinguish the integration variables and write the proper limits on all the integrals.