Is this integral undefined

Is this integral undefined

I have to evaluate the integral
$$\int_{0}^{\arcsin(g/r)}\left(\sum_{n=0}^{\infty}\sum_{m=0}^{n}\left(\arccos^{2n-2m}(x)(g^{2}-r^{2}+r^{2}\cos^{2}(x))^{m}\right)\right)\,dx$$
where $$a=\frac{(-1)^{n}(2n)!a^{n}r^{2n-2m}}{n!(2m)!(2n-2m)!},$$ which
comes out to $$\sum_{n=0}^{\infty}\sum_{m=0}^{n}\left(a(\phi)\right),$$
where
$$\phi=\left[\varphi(\cos^{2}(x))^{\frac{1}{2}+m-n}\sin(x)(g^{2}-r^{2}\sin^{2}(x))^{m}\left(1-\frac{r^{2}\sin^{2}(x)}{g^{2}}\right)^{-m}\right]_{0}^{\arcsin(g/r)}$$
and
$$\varphi=\operatorname{AppellF1}\left[\frac{1}{2},\frac{1}{2}+m-n,-m,\frac{3}{2},\sin^{2}(x),\frac{r^{2}\sin^{2}x}{g^{2}}\right]\cos^{-1-2m+2n}(x),$$
which all comes from Difficult Gaussian Integral Involving Two Trig
Functions in the Exponent: Any Help?
However, when I evaluate this at the upper limit I get terms in the sum
that involve
$$\left(1-\frac{r^{2}\sin^{2}(\arcsin(g/r))}{g^{2}}\right)^{-m}=(0)^{-m}$$
with $m$ climbing up so that we have terms in the integral that are
undefined ($0^{-a}$ is undefinied for positive $a$).
Does this mean the integral cannot be done between these limits? I'm
trying to plot the sum for various $g$ and $r$ but Matlab keeps giving me
NaN's due to the $0^{-1}$ terms.
Thanks.