Is there a way to represent $\frac{\int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx} $ in terms of a, b? We are asked to find the supremum of $\frac{exp(-x^2) \int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx}$ for $x \in (0, 1)$. I'm thinking since $\frac{\int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx}$ should be a fixed value, the supremum should just be $\frac{\int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx}$. But then I don't know how to simplify that.
I know that $\int_0^1 x^{a-1} (1-x)^{b-1} dx$ is just $B(a, b)$, and our instructor told us there is no need to calculate $\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx$, so I'm wondering is there anyway I can simplify that?

By expanding the exponential into a series and integrating then it will be found that
$$ \int_{0}^{1} e^{-x^2} \, x^{a-1} \, (1-x)^{b-1} \, dx = B(a,b) \, {}_{2}F_{2}\left(\frac{a}{2}, \frac{a+1}{2}; \, \frac{a+b}{2}, \frac{a+b+1}{2}; \, -1 \right). $$
This leads to
$$\frac{ e^{-x^2} \, \int_{0}^{1} x^{a-1} \, (1-x)^{b-1} \, dx}{\int_{0}^{1} e^{-x^2} \, x^{a-1} \, (1-x)^{b-1} \, dx} = \frac{e^{-x^2}}{{}_{2}F_{2}\left(\frac{a}{2}, \frac{a+1}{2}; \, \frac{a+b}{2}, \frac{a+b+1}{2}; \, -1 \right)}. $$
The hypergeometric function is a constant which gives the form
$$\frac{ e^{-x^2} \, \int_{0}^{1} x^{a-1} \, (1-x)^{b-1} \, dx}{\int_{0}^{1} e^{-x^2} \, x^{a-1} \, (1-x)^{b-1} \, dx} = \frac{1}{c_{0}} \, e^{-x^2}. $$
