Let $U_1, \dots, U_n$ be independent Uniform $[0, 1]$ random variables. Then it follows that the $k$th order statistic, $U_{(k)} \sim \text{Beta}(\alpha=k, \beta=n - k + 1)$ for $k = 1, \dots, n$.
Let $1 \leq i < j \leq n$ be two fixed integers. I wish to find $\mathbb{E}[U_{(i)}U_{(j)}]$.
The joint density of $(U_{(i)}, U_{(j)})$ is given by $$f_{U_{(i)}, U_{(j)}}(u, v) = \dfrac{n!}{(i-1)!(j-i-1)!(n-j)!}u^{i-1}(v-u)^{j-i-1}(1-v)^{n-j}\text{, } 0 \leq u < v \leq 1\text{.}$$ Thus, $$\mathbb{E}[U_{(i)}U_{(j)}] \propto\int_{0}^{1}\int_{0}^{v}u^{i}v(v-u)^{j-i-1}(1-v)^{n-j}\text{ d}u\text{ d}v\text{.}$$ (I'm ignoring the constant for now.) Perform the substitution $$w = \dfrac{u}{v}$$ to get $$\begin{align} \mathbb{E}[U_{(i)}U_{(j)}] &\propto\int_{0}^{1}\int_{0}^{1}(wv)^{i}v(v-wv)^{j-i-1}(1-v)^{n-j}\dfrac{\text{ d}w}{v}\text{ d}v \\ &= \int_{0}^{1}\int_{0}^{1}w^{i}v^{i+j-i-1}(1-w)^{j-i-1}(1-v)^{n-j}\text{ d}w\text{ d}v \\ &= \int_{0}^{1}\int_{0}^{1}w^{i}(1-w)^{j-i-1}v^{j-1}(1-v)^{n-j}\text{ d}w\text{ d}v \\ &= \int_{0}^{1}w^{i}(1-w)^{j-i-1}\text{ d}w \cdot \int_{0}^{1}v^{j-1}(1-v)^{n-j}\text{ d}v \\ &= B(i+1, j-i) \cdot B(j, n-j+1) \\ &= \dfrac{\Gamma(i+1)\Gamma(j-i)}{\Gamma(j+1)} \cdot \dfrac{\Gamma(j)\Gamma(n-j+1)}{\Gamma(n+1)}\tag{1} \end{align}$$ where $B(\cdot, \cdot)$ denotes the Beta function.
My professor has $\dfrac{\Gamma(i+1)\Gamma(j-i)}{\Gamma(j+1)} \cdot \dfrac{\Gamma(j+2)\Gamma(n-j+1)}{\Gamma(n+3)}$ where $(1)$ is, with the final result (including the constant) being $$\mathbb{E}[U_{(i)}U_{(j)}] = \dfrac{i(j+1)}{(n+2)(n+1)}\text{.}$$
Did I do something wrong?
This was a really stupid mistake. If $w = u/v$, then $\text{d}w = \text{d}u/v$, hence $\text{d}u = v\text{ d}w$, so $$\begin{align} \mathbb{E}[U_{(i)}U_{(j)}] &\propto\int_{0}^{1}\int_{0}^{1}(wv)^{i}v(v-wv)^{j-i-1}(1-v)^{n-j}v\text{ d}w\text{ d}v \\ &= \int_{0}^{1}\int_{0}^{1}w^{i}v^{i+j-i-1+2}(1-w)^{j-i-1}(1-v)^{n-j}\text{ d}w\text{ d}v \\ &= \int_{0}^{1}\int_{0}^{1}w^{i}(1-w)^{j-i-1}v^{j+1}(1-v)^{n-j}\text{ d}w\text{ d}v \\ &= \int_{0}^{1}w^{i}(1-w)^{j-i-1}\text{ d}w \cdot \int_{0}^{1}v^{j+1}(1-v)^{n-j}\text{ d}v \\ &= B(i+1, j-i) \cdot B(j+2, n-j+1) \\ &= \dfrac{\Gamma(i+1)\Gamma(j-i)}{\Gamma(j+1)} \cdot \dfrac{\Gamma(j+2)\Gamma(n-j+1)}{\Gamma(n+3)} \end{align}$$