Integral does not 'converge' despite describing a well-defined area.... I have almost evaluated (where all variables are real including the variable $i$)
$$
 C_1\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r \\ = \left[ \frac{1}{2} r^2 (i-r)^{\frac{3 (i-1)}{2}} \left(1-\frac{r}{i}\right)^{\frac{1}{2} (-3) (i-1)} \left(1-\frac{r}{i+j-R}\right)^{\frac{3}{2} (i+j-1)} \\(i+j-r-R)^{\frac{1}{2} (-3) (i+j-1)} F_1\left(2;\frac{1}{2} (-3) (i-1),\frac{3}{2} (i+j-1);3;\frac{r}{i},\frac{r}{i+j-R}\right) \right] _{a+bt^2}^{i}
$$
where $a+bt^2$ lies somewhere between $i-1$ and $i$. Also assume $i,j,r,R>0$; $F_1$ is the Appell hypergeometric series $F_1$.
There is a singularity at the upper limit (i.e. at $r=i$), despite the integral describing a well defined area (plotted here for $i=2,\hspace{1mm} j=2,\hspace{1mm}$ and $R=j-\frac{1}{5}$), so I can't continue...

Can anyone explain how I'm meant to perform this definite integral if there is a singularity at $r=i$?
 A: To answer my own question, after looking at Convergence of the improper integral $\int_{0}^{\pi/2}\tan^{p}(x) \; dx$ I tried the substitution $u = i-r$ and got 
$$
\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r =\\\frac{2\left(-a-b t^2+i\right)^{\frac{3 i}{2}-\frac{1}{2}}}{\left(9 i^2-1\right) \left(-\frac{a+b t^2-i-j+R}{j-R}\right)^{3/2}}  \\ \left(\left(\frac{1}{-a-b t^2+i+j-R}\right)^{\frac{3}{2} (i+j-1)} \left(1-\frac{a+b t^2-i}{j-R}\right)^{\frac{3 (i+j)}{2}} \\ \left(i (3 i+1) \, _2F_1\left(\frac{1}{2} (3 i-1),\frac{3}{2} (i+j-1);\frac{1}{2} (3 i+1);\frac{b t^2+a-i}{j-R}\right) \\ +(3 i-1) \left(a+b t^2-i\right) \, _2F_1\left(\frac{1}{2} (3 i+1),\frac{3}{2} (i+j-1);\frac{3 (i+1)}{2};\frac{b t^2+a-i}{j-R}\right)\right)\right)
$$
since the upper limit is now $0$ and makes the integral proper. Using a limit
$$
\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r =\lim_{W \to i}\int_{a + bt^2}^{W} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r
$$
may also work (see http://en.wikipedia.org/wiki/Improper_integral), though would be tricky due to the complexity of the evaluated definite integral between $a+bt^2$ and $W$ (if it even exists without using a substitution).
