TITLE: Upper and lower bound on integral
QUESTION [6 upvotes]: Consider the following integral
$$\int_0^1 (1-x^n)^M \,d x$$
It converges to $0$ as $M\to\infty$, but I would like to find bounds on the convergence rates. What I mean is that it is straightforward to find constants A and B such that 
$$\frac{A}{M}<\int_0^1 (1-x^n)^M \,d x<\frac{B}{M^{1/n}}$$
However, is it possible to obtain identical upper and lower bounds? The claim that I have is that the lower bound can also be made of the form $A'/M^{1/n}$, but I have not been able to prove it.

REPLY [2 votes]: After evaluating your integral as a beta function as in the beginning of draks' post, you can use Stirling's approximation for the gamma funcion to find an asymptotic series (for another approach see this paper).  Truncating at the third term gives
$$\int_0^1 (1-u^n)^M \,du = \frac{\Gamma(1+1/n)}{M^{1/n}}\left\{1 - \frac{n+1}{2n^2M} + O\left(\frac{1}{M^2}\right)\right\} \hspace{1cm} (M \to \infty).$$