Trying to find bounds on this integral: I am trying to find a bound on this integral: $\int_x^t(u-x)^{H-3/2}u^{H-1/2}du$ where $t> x$ $0\le H\le1$ and $0\le x\le1$, but the kicker is that I need the bound to not depend on $x$. I am new to this kind of stuff, and I have tried a few tricks I know but they always seem to end up with the integral being undefined. It is important to notice that $H-3/2<0$, which is something that causes the problem to be tricky. Also the $x$ inside the integrand also makes the problem be tricky. It may not be possible as well, I am not sure.
 A: Since you integral is not convergent for $0\leq H\leq 1/2$, I will below assume that $1/2<H\leq 1$. So, let $t>x$, with $0\leq x\leq 1$. The bound I give below will depend on $t$. I hope that is allowed (you don't say anything else). Also, it will blow up as $H\to 1/2^+$. I don't claim any optimality, but only give a bound independent of $x$.
New solution
As suggested by @JanG, it is easier to just integrate by parts:
$$
\begin{aligned}
\int_x^t (u-x)^{H-3/2}u^{H-1/2}\,du&=\frac{1}{H-1/2}(t-x)^{H-1/2}t^{H-1/2}-\int_x^t(u-x)^{H-1/2}u^{H-3/2}\,du\\
&\leq \frac{1}{H-1/2}(t-0)^{H-1/2}t^{H-1/2}-0\\
&=\frac{1}{H-1/2}t^{2H-1}.
\end{aligned}
$$
Old solution
As a start, we note that $u\mapsto (u-x)^{H-3/2}$ is decreasing in $(x,t)$ and that $u\mapsto u^{H-1/2}$ is increasing in $(x,t)$. Hence, (I've discussed this here before)
$$
\int_x^t (u-x)^{H-3/2}u^{H-1/2}\,du\leq \frac{1}{t-x}\int_x^t(u-x)^{H-3/2}\,du\int_x^t u^{H-1/2}\,du
$$
The first integral is calculated to be
$$
\int_x^t(u-x)^{H-3/2}\,du=\frac{1}{H-1/2}(t-x)^{H-1/2}.
$$
Inserting this, we find that
$$
\int_x^t (u-x)^{H-3/2}u^{H-1/2}\,du\leq\frac{1}{H-1/2}\frac{(t-x)^H}{(t-x)^{3/2}}\int_x^t u^{H-1/2}\,du.
$$
The integral in the right-hand side can be bounded (just estimate with the maximal value of the integrand, which occurs when $u=t$)
$$
\int_x^t u^{H-1/2}\,du\leq (t-x)t^{H-1/2}.
$$
Hence,
$$
\int_x^t (u-x)^{H-3/2}u^{H-1/2}\,du\leq\frac{1}{H-1/2}\frac{(t-x)^H}{(t-x)^{1/2}}t^{H-1/2}=\frac{1}{H-1/2}(t-x)^{H-1/2}t^{H-1/2}
$$
But $x\mapsto (t-x)^{H-1/2}$ is decreasing, so it attains it largest value $t^{H-1/2}$ for $x=0$. Inserting this bound, we end up with

$$
\int_x^t (u-x)^{H-3/2}u^{H-1/2}\,du\leq \frac{1}{H-1/2}t^{2H-1}
$$ 

Comment
If you would like to improve, one way could be to calculate both integrals after the first inequality to get that your integral is bounded by
$$
\frac{1}{H^2-1/4}\frac{t^{H+1/2}-x^{H+1/2}}{(t-x)^{H-3/2}}
$$
Below is a plot in the particular case $t=3/2$ and $H=3/4$. The green graph represents the true integral, the blue the one from the first integral estimate (i.e. the function in the comment above) and the orange the uniform estimate in $x$. 

