# Calculate $\lim_{n\rightarrow\infty} \int_0^1 nx^2(1-x^2)^n \, dx$

I have to calculate $$\lim_{n\rightarrow\infty} \int_0^1 nx^2(1-x^2)^n \, dx$$

I've created the series $(f_n)_{n\in \mathbb{N}}$ with $f_n:[0,1]\rightarrow \mathbb{R}, f_n(x)=nx^2(1-x^2)^n$. I considered $x\in[0,1]$ a scalar and proceed to calculated the limit $\lim_{n\rightarrow\infty}f_n(x)$, which equals to $0, \forall x\in[0,1)$ (if I'm not wrong), but I'm getting stuck at calculating the limit for the particular case of $x=1$. Thanks for help!

• Isn't $f_n(1)=0$ for all $n$? Commented Jun 21, 2017 at 19:01
• @uniquesolution I don't get if that is a rethorical question, but my guess is not. I think that if $x=1$ and $n\rightarrow\infty$, then $f_n\rightarrow \infty\cdot 0^{\infty}$ Commented Jun 21, 2017 at 19:04
• I think that if $x=1$ then $(1-x^2)=0$, hence $f_n(1)=0$. Commented Jun 21, 2017 at 19:05
• @uniquesolution so you need not compute the limit? Commented Jun 21, 2017 at 19:07
• I do, but it is spectacularly easy, because in the special case of $x=1$ the sequence is constant, and equals to zero. Commented Jun 21, 2017 at 19:08

Alternative approach: by setting $x=\sin\theta$ we have

$$I(n) = \int_{0}^{1}nx^2(1-x^2)^n\,dx = n\int_{0}^{\pi/2}\sin^2(\theta)\cos^{2n+1}(\theta)\,d\theta \tag{1}$$ and since over the interval $\left[0,\frac{\pi}{2}\right]$ both $\sin(\theta)$ and $\cos(\theta)$ are non-negative, but $\sin(\theta)\leq\theta$ and $\cos(\theta)\leq e^{-\theta^2/2}$, it follows that: $$0\leq I(n) \leq \int_{0}^{\pi/2} n\theta^2 e^{-\left(n+\frac{1}{2}\right)\theta^2}\,d\theta \leq \int_{0}^{+\infty} n\theta^2 e^{-\left(n+\frac{1}{2}\right)\theta^2}\,d\theta=\sqrt{\frac{\pi}{2}}\frac{n}{(2n+1)^{3/2}}\tag{2}$$ and the wanted limit is zero by squeezing.

• why not applying $1-x<e^{-x}$ directly? :) Commented Jun 21, 2017 at 19:31
• @tired: because this approach (Laplace's method in disguise) leads to an almost-optimal bound, and I am partial to very accurate approximations :) Commented Jun 21, 2017 at 19:57
• @JackD'Aurizio Then why not use the approach I used, which evaluates the integral in closed form and uses Stirling's Formula to show that the integral is of order $n^{-1/2}$? Commented Jun 30, 2017 at 22:18
• @MarkViola: it works, of course, but I prefer to use this approach (essentially, Laplace's method) or creative telescoping to prove Stirling's formula, so in my mind this technique is a bit more "elementary" than invoking Stirling's formula. Just a matter of taste. Commented Jul 1, 2017 at 0:37

Since $1-x^2<e^{-x^2}$ for each $0\leq x\leq 1$, the integrand is non-negative and bounded above by $nx^2e^{-nx^2}$. Since $ye^{-y}$ is a bounded function on $[0,\infty)$, and $nx^2e^{-nx^2}\to 0$ for all $x\geq 0$, the result follows from the dominated convergence theorem.

Integrating by parts, $$\int_0^1 nx^2(1-x^2)^n \, dx = \left[ -\frac{n}{2(n+1)}x(1-x^2)^{n+1} \right]_0^1 + \frac{n}{2(n+1)}\int_0^1 x(1-x^2)^{n+1} \, dx \\ = \frac{n}{2(n+1)}\int_0^1 (1-x^2)^{n+1} \, dx$$ The fraction converges to $1/2$, so now we need to look at the remaining integral. $(1-x^2)^{n+2} < (1-x^2)^{n+1}$, so the integrand decreases as $n \to \infty$. We can then use the monotone convergence theorem or similar to show the integral tends to zero, so the limit is zero.

• nice$\mathbb{+1}$! Commented Jun 21, 2017 at 19:14
• Why not use the dominated convergence theorem and spare the integration by parts? Commented Jun 21, 2017 at 19:15
• @uniquesolution Is there an obvious dominating function? Commented Jun 21, 2017 at 19:17
• @Chappers - you are right, there is no obvious dominating function. Commented Jun 21, 2017 at 19:24
• @Chappers but if you note that $(1-x^2)\leq e^{-x^2}$ for $0\leq x\leq 1$, and then note that $ye^{-y}$ is a bounded function, you can use the dominated convergence theorem directly. Commented Jun 21, 2017 at 19:37

The integral can be evaluated in closed form before letting $n\to \infty$. To that end we now proceed.

Enforcing the substitution $x\to \sqrt{x}$, we find

\begin{align} \int_0^1 nx^2(1-x^2)^n\,dx&=\frac n2\int_0^1 x^{1/2}(1-x)^n\,dx\\\\ &=\frac n2 B\left(\frac32,n+1\right)\\\\ &=\frac n2 \frac{\Gamma(3/2)\Gamma(n+1)}{\Gamma(n+5/2)}\\\\ &=\frac n2 \frac{\color{blue}{\Gamma(3/2)}\color{orange}{\Gamma(n+1)}}{(n+3/2)(n+1/2)\color{red}{\Gamma(n+1/2)}}\\\\ &=\frac n2 \frac{\color{blue}{\frac{\sqrt\pi}{2}}\color{orange}{(n!)}}{(n+3/2)(n+1/2)\color{red}{\left(\frac{2^{1-2n}\sqrt\pi\,\Gamma(2n)}{\Gamma(n)}\right)}}\\\\ \therefore \int_0^1 nx^2(1-x^2)^n\,dx&=\frac{4^n\,n\,(n!)^2}{(2n+3)(2n+1)(2n)!}\\\\ &=O\left(\frac{1}{\sqrt n}\right)\,\,\,\,\dots \text{Applying Stirling's Formula} \end{align}

Hence, as $n\to \infty$, the integral of interest goes to $0$ as $\frac1{\sqrt n}$.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\int_{0}^{1}nx^{2}\pars{1 -x^{2}}^n\,\dd x & = \lim_{n \to \infty}\bracks{n% \int_{0}^{1}\exp\pars{2\ln\pars{x} + n\ln\pars{1 - x^{2}}}\,\dd x} \end{align}

The $\ds{\exp}$-argument has a 'sharp maximum' at $\ds{x_{n} = \pars{n + 1}^{-1/2}}$ such that

\begin{align} \lim_{n \to \infty}\int_{0}^{1}nx^{2}\pars{1 -x^{2}}^n\,\dd x & = \lim_{n \to \infty}\bracks{\pars{n \over n + 1}^{n + 1}% \int_{0}^{\infty}\exp\pars{-\,{\bracks{x - x_{n}}^{\,2} \over 2\sigma_{n}^{2}}}\,\dd x} \end{align}

where $\ds{\sigma_{n} \equiv {\root{n} \over 2\pars{n + 1}}}$.

Then, \begin{align} \lim_{n \to \infty}\int_{0}^{1}nx^{2}\pars{1 -x^{2}}^n\,\dd x & = \expo{-1}\root{\pi \over 2}\lim_{n \to \infty}\braces{% \sigma_{n}\bracks{1 + \mrm{erf}\pars{x_{n} \over \root{2}\sigma_{n}}}} \\[5mm] & = \expo{-1}\root{\pi \over 2}\bracks{1 + \mrm{erf}\pars{\root{2}}} \lim_{n \to \infty}\sigma_{n} = \bbx{0} \end{align}