I was supposed to prove that $\int_{-1}^1 \frac{f(x)g(x)}{\sqrt(1-x^2)} dx$ converges.
However, I am not sure if this actually converges because I was not given any information about $f(x)$ and $g(x)$ other than that $f(x), g(x) \in \mathbb{R}[x]$. Could someone give any insights? Thanks
Are $f$ and $g$ continuous? If they are, then the only possibility of non-convergence is at the endpoints $\pm1$ of the interval. So we ask if the integrals over $[1-\epsilon,1]$ and $[-1,-1+\epsilon]$ converge. These are similar, so look at the second. A change of variables gives $$\int_{-1}^{-1+\epsilon}\frac{h(x)\,dx}{\sqrt{1-x^2}} =\int_0^\epsilon\frac{h(y-1)\,dy}{\sqrt{y(2-y)}}=\int_0^\epsilon\frac {k(y)\,dy}{\sqrt y}$$ for suitable continuous functions $h$ and $k$. As $k$ is continuous, it is bounded on $[0,\epsilon]$, so the integral is bounded by a constant multiple of $\int_0^\epsilon \frac{dy}{\sqrt y}$ which is a convergent integral.