I would like to bound the following integral with an upper bound depending on $a\in[0,1]$ when $a$ is close to 1:
$$\int_a^1\frac{t^{\frac{r-3}{2}}}{\sqrt{1-t}}dt$$ where $r\ge 5$ is an integrer.
I have tried integration by parts with some steps. I also computed numerically, it seems that the dependance with $a$ is exponential close to 1.
If you invoke the Gausssian hypergeometric function, assuming that I am not mistaken, the integral is $$I=\sqrt{\pi } \,\frac{\Gamma \left(\frac{r-1}{2}\right)}{\Gamma \left(\frac{r}{2}\right)}-\frac{2 a^{\frac{r-1}{2}} }{r -1}\,\ _2F_1\left(\frac{1}{2},\frac{r-1}{2};\frac{r+1}{2};a\right)$$
Expanded as series around $a=1^-$ $$\color{blue}{ I=t-\frac{r-3}{24} t^3+\frac{(r-5) (r-3)}{640} t^5+O\left(t^7\right)}\quad \text{with} \quad \color{blue}{ t=2 \sqrt{1-a}}$$
Trying for $r=8$ and $a=\frac9{10}$ this gives $$I=\frac{2209}{1200 \sqrt{10}}=0.582123$$ while the exact value is $$I=\frac{381}{1000}+\frac{5 \pi }{16}-\frac{5}{8} \tan ^{-1}(3)=0.582094$$
This does not look as an exponential behaviour.
If you are concerned by approximations of $I$, we could have much better using more terms in the series to be later transformed into a $[n+1,n]$ Padé approximant $P_n$. For the worked example, $P_4$ gives an absolute error of $2.21\times 10^{-8}$.