I am trying to solve the following integral using the steepest descent method:

$$I\_n(\alpha)=\int\_0^1x^{\alpha n} (1-x)^{\beta n} dx$$

$\beta=1-\alpha , \alpha > 0, \beta > 0$.

This is from a book, and the solution says that the saddle point is at $x=\alpha$. I cannot see why. I tried to put it in the form: $$\int\_a^be^\frac{-f(x)}{\lambda}dx$$

To do this, I used $x^{\alpha n}=e^{\alpha n \ln(x)}$ and $(1-x)^{\beta n} = e^{\beta n\ln(1-x)}$. After simplifying, this gave me $\lambda=\frac{1}{n}$ and $f(x)=\ln(x^\alpha(1-x)^{1-\alpha})$. To minimise $f(x)$, I don't see how $x$ can equal $\alpha$. Unfortunately, the book doesn't give more information on how the saddle point is at $x=\alpha$. I would like to know how to obtain this result and complete the integral.