Integral of $-\alpha\int_{x_l}^1F(x)dF(x)$?, $F(x)$ continuous with support $[x_l,1]$ where $x_l$ such that $supp(F(x)) =[x_l,1]$ (supp means support)
Also, $F(x)$ is continuous
I don't know how to deal with the $F(x)$ in the integral.

From the first line we know that $F(x_l)=0$ and $F(1)=1$, so if there is some way to write out the integral in terms of those two things then we know the answer. But I can't think of a way to write is a such (without there being a constant?), and I can't think of another approach.
 A: $$-\alpha\int_{x_l}^1F(x)dF(x)$$
$$ = -\alpha \frac{F(x)^2}{2}|_{x_l}^{1}$$
$$ = - \frac{\alpha}{2} F(x)^2|_{x_l}^{1}$$
$$ = - \frac{\alpha}{2} [(1)^2 - (x_l)^2]$$
A: In the sense of Riemann-Stieltjes integral, we can provide a direct proof:

Claim. If $f : [a, b] \to \mathbb{R}$ is continuous and monotone increasing, then
  $$ \int_{a}^{b} f(x) \, df(x) = \left[ \frac{f(x)^2}{2} \right]_{a}^{b} $$
  as in Riemann-Stieltjes integral sense.

Indeed, let $\Pi = \{a = x_0 < \cdots < x_n = b \}$ be any partition of $[a, b]$. Then the upper sum $U(\Pi, f)$ and the lower sum $L(\Pi, f)$ is given by
\begin{align*}
U(\Pi, f) &= \sum_{i=1}^{n} f(x_i)[f(x_i) - f(x_{i-1})], \\
L(\Pi, f) &= \sum_{i=1}^{n} f(x_{i-1})[f(x_i) - f(x_{i-1})].
\end{align*}
Now notice that
\begin{align*}
U(\Pi, f) - L(\Pi, f)
&= \sum_{i=1}^{n} [f(x_i) - f(x_{i-1})]^2 \\
&\leq [f(b) - f(a)]\max_{1\leq i\leq n} [f(x_i) - f(x_{i-1})]
 \xrightarrow[\|\Pi\|\to0]{} 0
\end{align*}
by the uniform continuity of $f$ on $[a, b]$ and that
$$ U(\Pi, f) + L(\Pi, f) = \sum_{i=1}^{n} [f(x_i)^2 - f(x_{i-1})^2] = \left[ f(x)^2 \right]_{a}^{b}. $$
Therefore the desired Riemann-Stieltjes integral exists with value $\frac{1}{2}[f(b)^2 - f(a)^2]$.
One can prove a more general statement in a similar spirit:

Proposition. (Integration by parts) If $f, g$ are functions on $[a, b]$ such that $\int_{a}^{b}f(x) \, dg(x)$ exists as Riemann-Stieltjes integral, then $\int_{a}^{b}g(x) \, df(x)$ also exists and
  $$ \int_{a}^{b}f(x) \, dg(x) = [f(x)g(x)]_{a}^{b} - \int_{a}^{b}g(x) \, df(x). $$


An overkill. In the sense of Lebesgue-Stieltjes integral, we can provide another proof.

Claim. If $f : [a, b] \to \mathbb{R}$ is right-continuous and monotone increasing, then
  $$ \int_{a}^{b} f(x) \, df(x) = \left[ \frac{f(x)^2}{2} \right]_{a}^{b} + \frac{1}{2}\sum_{x \in (a,b]} [f(x) - f(x^-)]^2, $$
  as in Lebesgue-Stieltjes integral sense, where $f(x^-) = \lim_{y \uparrow x} f(y)$ is the left-limit and hence $f(x) - f(x^-)$ is the jump size of $f$ at $x$.

Indeed, there exists a Stieltjes measure $\mu$ such that $f(x) = \mu([a,x])$ and hence
\begin{align*}
2\int_{a}^{b} f(x) \, df(x)
&= \int_{[a,b]} \int_{[a,x]} \mu(dy)\mu(dx) + \int_{[a,b]} \int_{[a,y]} \mu(dx)\mu(dy) \\
&= \int \left( \mathbf{1}_{\{ a \leq y \leq x \leq b\}} + \mathbf{1}_{\{ a \leq x \leq y \leq b\}} \right) \, \mu^{\otimes 2}(dx,dy) \\
&= \int \left( \mathbf{1}_{[a,b]^2} + \mathbf{1}_{\{ a \leq y = x \leq b\}} \right) \, \mu^{\otimes 2}(dx,dy) \\
&= \mu([a,b])^2 + \int \left( \sum_{x \in [a, b]} \mu(\{x\}) \mathbf{1}_{\{y=x\}} \right) \, \mu(dy) \\
&= \mu([a,b])^2 + \sum_{x\in[a,b]} \mu(\{x\})^2.
\end{align*}
Rewriting everything in terms of $f$ yields the desired equality.
