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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.

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Where you wrote "as $m\to\infty$", did you mean "as $M\to\infty$"? – Michael Hardy Feb 16 '12 at 21:24
Yes, thank you! – Ivan Feb 16 '12 at 21:39
Would a closed solution also be fine? – draks ... Feb 16 '12 at 21:48
I am not sure I understand what you mean by closed solution. I could expand by the binomial formula and integrate term by term, but I am not able to figure out the leading term in the resulting expansion. – Ivan Feb 16 '12 at 21:56
up vote 2 down vote accepted

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


(Assuming I have done the calculations right).

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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 $T(I_M)= _2F_1(1,(M-1);\frac{1}{n}+k-1;1)$

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Wolfram Alpha is unable to evaluate this. Can you show we get that? +1 in anticipation. – Aryabhata Feb 17 '12 at 3:38
But it did, so you say it wrong? – draks ... Feb 17 '12 at 8:07
No, I am not saying it is wrong. All I was requesting was to expand on it a bit more! – Aryabhata Feb 17 '12 at 17:08
@Aryabhata I expanded as far as I could (for now ;-)! – draks ... Feb 22 '12 at 20:26
@draks: I have edited your comment to include a derivation of your first equality, but I don't yet have free edit privileges so a mod will need to approve it before it posts. – Antonio Vargas Feb 22 '12 at 21:10

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

$$\int_0^1 (1-u^n)^M \,du = \frac{\Gamma(1+1/n)}{M^{1/n}}\left\{1 - \frac{n+1}{2n^2M} + O\left(\frac{1}{M^2}\right)\right\} \hspace{1cm} (M \to \infty).$$

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