I have two expressions (let's call them functions $f,g$) on $[0,1]$, where I want to find out whether they are square-integrable or better: for which $m \in \mathbb{Z}$ they are square-integrable ( the problem of course is that it is impossible to evaluate the integrals explicitely):
$$f(x) = (1-x^2)^{\frac{1}{4}} \frac{\sqrt{1+x}^m}{\sqrt{1-x}^m} e^{ \beta x} \int_0^x \frac{\sqrt{\frac{1-t}{1+t}}^{2m}}{(1-t^2)e^{2 \beta t}} dt$$
and
$$g(x) = (1-x^2)^{\frac{1}{2}-m} e^{ \beta x} \int_0^x \frac{1}{\sqrt{1-t^2} \sqrt{1-t^2}^{1-2m}e^{2 \beta t}} dt.$$
The only thing I know for sure is that the prefactors are at least square integrable for $m \le 0$.
Let's look at $f$. In what follows, $C_i$ are functions which are bounded away from both $0$ and $\infty$. The prefactor may be written as
$$C_1(x) (1-x)^{-m/2+1/4}$$
The integrand may be written as
$$C_2(t) (1-t)^{m-1}$$
Thus for $m \neq 0$, the integral may be written as
$$C_3(x) (1-x)^m$$
while for $m=0$ we have
$$C_4(x) \ln(1-x)$$
Thus overall for $m \neq 0$ we have
$$f(x) = C_5(x) (1-x)^{m/2+1/4}$$
and for $m = 0$ we have
$$f(x) = C_6(x) (1-x)^{1/4} \ln(1-x)$$
The latter is bounded, so it is square integrable. The former, when squared, looks like
$$f^2(x) = C_7(x) (1-x)^{m+1/2}$$
which is integrable provided $m+1/2>-1$. So $f$ is square integrable provided $m \geq -1$ and is not square integrable if $m < -1$.
$g$ can be handled in essentially the same fashion as $f$.