Upper and lower bound on integral 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.
 A: Assuming $n \gt 1$,
If we set $I_{M} = \int_{0}^{1} (1-x^n)^M dx$, then, I believe we get, using integration by parts ($u = x$, $v = (1-x^n)^M$) that
$$I_{M+1} = \int_{0}^{1} (M+1)n x^n (1-x^n)^M dx$$
and so
$$ (M+1)n I_{M} - I_{M+1} = (M+1)n \int_{0}^{1} (1-x^n)^{M+1} dx = (M+1)nI_{M+1}$$
$$I_{M+1} = \frac{(M+1)n}{(M+1)n +1}I_{M} = \left(1 - \frac{1}{(M+1)n + 1}\right) I_m$$
Now we can use the estimate $1 - \frac{1}{x} = e^{-x + O(x^2)}$ and get an estimate for $I_{M}$.
All we would need is an estimate for $\sum_{k=0}^{M} \frac{1}{kn+1}$ which I believe is $\frac{\log M}{n} + O(\frac{1}{M})$ and thus your integral is
$$\Theta\left(\frac{1}{M^{1/n}}\right)$$
(Assuming I have done the calculations right).
A: We start with some expressions for the beta function,
$$
B(p,q) = 2 \int_0^1 x^{2p-1} (1-x^2)^{q-1} \,dx = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}
$$
(see e.g. equation (21) here), where $\Gamma$ is the gamma function.  Setting $p=\frac{1}{n}$ and $q=M+1$, multiplying by $\frac{1}{n}$, and using the identity $\Gamma(x+1) = x\Gamma(x)$ we get
$$
\frac{2}{n} \int_0^1 x^{\frac{2}{n}-1} (1-x^2)^{M} \,dx = \frac{\Gamma(M+1)\Gamma(1+1/n)}{\Gamma(M+1+1/n)}.
$$
Making the substitution $x = u^{n/2}$ in the integral gives your integral on the left-hand side.  That is,
$$
\int_0^1 (1-u^n)^M \,du = \frac{\Gamma(M+1)\Gamma(1+1/n)}{\Gamma(M+1+1/n)}.
$$
The right-hand side can be rewritten as
$$
I_M=\frac{\Gamma(M+1)\Gamma(1+1/n)}{\;\Gamma(M+1+1/n)}=\frac{M! \; (1/n)! }{(M+1/n)!}=
\binom{M+1/n}{M}^{-1}.
$$
So why is that? Let's start with exanding $(1-x^n)^M$. We get
$$
I_M=\int_0^1 \sum_{k=0}^M (-1)^k \binom{M}{k} x^{nk} dx= \sum_{k=0}^M (-1)^k \frac{\binom{M}{k}}{nk+1}. $$
The last sum may be recognized as Binomial Transform $T$ of the sequence $\{ \frac{1}{nk+1} \}$. Since $TT=1$, we transform the result to
$$
T(I_M)=\sum_{k=0}^M (-1)^k \frac{\binom{M}{k}}{\binom{k+1/n}{k}}=\sum_{k=0}^M (-1)^k \frac{M!\frac{1}{n}!}{(M-k)!(k+\frac{1}{n})!},
$$
which gives $\displaystyle T(I_M)= \frac{1}{Mn+1}$.
The heavy use of WolframAlpha makes me believe, that what I showed is not enough.
Answers to this question would give me better feeling.
Maybe this could help there: The last sum could be rewritten using Pochhammer Symbols as
$$
T(I_M)=\sum_{k=0}^M  \frac{(M-1)_{k}(1)_k}{(\frac{1}{n})^{(k)}}\frac{(-1)^k}{k!}=\sum_{k=0}^M  \frac{(M-1)_{k}(1)_k}{(\frac{1}{n}+k-1)_{(k)}}\frac{(-1)^k}{k!}
$$
and if $M\to \infty$ we get $T(I_M)= _2F_1(1,(M-1);\frac{1}{n}+k-1;1)$
A: 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
$$\int_0^1 (1-u^n)^M \,du = \frac{\Gamma(1+1/n)}{M^{1/n}}\left\{1 - \frac{n+1}{2n^2M} + O\left(\frac{1}{M^2}\right)\right\} \hspace{1cm} (M \to \infty).$$
