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I'd like your help with the following claim to prove:

$$\lim_{n \to \infty} \int_{0}^{\sqrt n}\left(1-\frac{x^2}{n}\right)^ndx=\int_{0}^{\infty} e^{-x^2}dx.$$

I think I should use the claim:

Let $a,b$ two real numbers and $\{f_n\}$ a sequence of continuous functions on $\left[a,b\right]$ which converges uniformly to $f$ on $[a,b]$. Then $$\lim_{n\to\infty}\int_a^bf_n(t)dt=\int_a^bf(t)dt.$$

But for this, I must prove that $(1-\frac{x^2}{n})^n$ uniformly converges to $e^{-x^2}$. How do I do that? One of the hardest and trickiest things to do is to prove this uniformly converges.. every function has its own way. I believe that the more examples I'll see the easiest it will be. Thanks again!

pie
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Jozef
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6 Answers6

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Although perhaps not the best approach here, I give a proof with bare hands. The idea is to bound the integrand $f(x) = \left( 1- \frac{x^2}{n} \right)^n$, and hence the integral, directly.

  • Upper bound. Since $1 - z \leqslant \mathrm e^{-z}$ for all $z$, we have we have $f(x) \leqslant \mathrm e^{-x^2}$ for $0 \leqslant x \leqslant \sqrt{n}$. Therefore, $$ \int_{0}^{\sqrt{n}} f(x) \, dx \leqslant \int_{0}^{\sqrt{n}} \mathrm e^{-x^2} \, dx \lt \int_{0}^{\infty} \mathrm e^{-x^2} \, dx. $$

  • Lower bound. Here we need the standard estimate $1 -z \geqslant \mathrm e^{-z - z^2}$ valid for $0 \leqslant z \leqslant \frac14$. (This can be obtained by using Taylor's theorem for $\log (1-z)$ at $z=0$.) Therefore, it follows that for $0 \leqslant x \leqslant \frac12 \sqrt{n}$, $$f(x) \geqslant \exp \left( - x^2 - \frac{x^4}{n} \right) .$$ In particular, for $0 \leqslant x \leqslant n^{1/8} \lt \frac12 \sqrt{n}$ (for large enough $n$), we have $ f(x) \geqslant \mathrm{e}^{-x^2} \mathrm e^{- \frac{1}{\sqrt{n}}} $.

    Therefore, $$ \begin{align*} \int_0^{\sqrt{n}} f(x) \, dx &\geqslant \int_0^{n^{1/8}} f(x) \, dx \\ &\geqslant \int_0^{n^{1/8}} \mathrm{e}^{-x^2} \mathrm e^{- \frac{1}{\sqrt{n}}} \, dx \\ &= \mathrm e^{-\frac{1}{\sqrt{n}}} \cdot \int_0^{n^{1/8}} \mathrm{e}^{-x^2}\, dx \end{align*} $$ Notice that the sequence $\mathrm e^{-\frac{1}{\sqrt{n}}} \cdot \int_0^{n^{1/8}} \mathrm{e}^{-x^2}\, dx$ converges to $\int_0^\infty \mathrm e^{-x^2} \, d x$.

Therefore by the sandwich theorem, we can conclude that $$ \int_0^{\sqrt{n}} f(x) \, dx \to \int_0^\infty \mathrm e^{-x^2} \, dx. $$

Srivatsan
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    Indeed, $1-z\ge e^{-z-z^2}$ is valid on $[0,1/2]$, at least. First, consider where the derivative of $z+z^2+\log(1-z)$ is $0$: $$1+2z-\frac{1}{1-z}=0\Leftrightarrow z-2z^2=0\Leftrightarrow z\in{0,1/2}$$ Thus, $z+z^2+\log(1-z)$ is monotonic on $[0,1/2]$ and non-negative at both ends. Thus, $\log(1-z)\ge-z-z^2$ and therefore, $1-z\ge e^{-z-z^2}$ – robjohn Dec 15 '11 at 08:25
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I fail to see how the behaviour of the integrals on $[a,b]$ could yield the $[0,+\infty)$ case.

Hint for a solution: consider the functions $f$ and $f_n$ defined on $\mathbb R_+$ by $f(x)=\mathrm e^{-x^2}$ and $$ f_n(x)=\left(1-\frac{x^2}n\right)^n\cdot[x\leqslant\sqrt{n}]. $$ Show that the sequence $(f_n)_{n\geqslant1}$ is increasing and that $f_n\to f$ pointwise. Then, find a theorem in your notes which guarantees that, in this setting, the integral of $f_n$ converges to the integral of $f$.

Did
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$$\int_{0}^{\sqrt{n}}\left(1-\frac{x^2}{n}\right)^n\,dx \stackrel{x\mapsto\sqrt{z}}{=}\frac{1}{2}\int_{0}^{n}z^{-1/2}\left(1-\frac{z}{n}\right)^n\,dz \stackrel{z\mapsto nu}{=}\frac{\sqrt{n}}{2}\int_{0}^{1} u^{-1/2}(1-u)^n\,du $$ The last integral is a value of Euler's Beta function. By integration by parts we get: $$ \frac{\sqrt{n}}{2}\cdot \frac{\Gamma\left(\frac{1}{2}\right)\,\Gamma(n+1)}{\Gamma\left(n+\frac{3}{2}\right)} $$ and by Gautschi's inequality we have:

$$ \lim_{n\to +\infty}\frac{\sqrt{n}}{2}\cdot \frac{\Gamma\left(\frac{1}{2}\right)\,\Gamma(n+1)}{\Gamma\left(n+\frac{3}{2}\right)} = \frac{1}{2}\,\Gamma\left(\frac{1}{2}\right) = \color{blue}{\frac{1}{2}\sqrt{\pi}}.$$

With a more elementary approach, we may just approximate $\left(1-\frac{x^2}{n}\right)^n$ over $(0,\sqrt{n})$ with $\left(e^{-x^2/n}\right)^n = e^{-x^2}$ in the spirit of Laplace's method. Since over $(0,1)$ we have $0\leq e^{-x^2}-(1-x^2)\leq \frac{x^3}{e}$ and $0<a<b$ grants $b^n-a^n\leq n(b-a)b^{n-1}$,

$$ I_n = \int_{0}^{\sqrt{n}}\left(1-\frac{x^2}{n}\right)^n\,dx = \int_{0}^{\sqrt{n}}e^{-x^2}\,dx - \frac{C}{n^2}\int_{0}^{\sqrt{n}}x^3 e^{-x^2}\,dx $$ with $C\in[0,1]$. On the other hand $$ \int_{\sqrt{n}}^{+\infty}e^{-x^2}\,dx = \int_{n}^{+\infty}\frac{e^{-z}}{2\sqrt{z}}\,dz\leq \frac{1}{2\sqrt{n}}\int_{n}^{+\infty}e^{-z}\,dz = \frac{1}{2e^n\sqrt{n}}$$ hence: $$ I_n = O\left(\frac{1}{n^2}\right) + \int_{0}^{+\infty}e^{-x^2}\,dx = \color{blue}{\frac{\sqrt{\pi}}{2}}+O\left(\frac{1}{n^2}\right).$$

Community
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Jack D'Aurizio
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Hint : Let $f_n(x)=\left( {1 - \frac{{{x^2}}}{n}} \right)^n\chi_{[0,\sqrt{n}]}$ and use Monotone convergence theorem.

Bérénice
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If you want to use uniform convergence, you can do the following. Since $\ln(1 - y) \leq -y$ for $0 \leq y < 1$ (look at the power series), you have $n\ln(1 - {x^2 \over n}) \leq -n{x^2 \over n} = -x^2$, and taking exponentials you get $(1 - {x^2 \over n})^n \leq e^{-x^2}$ on your range of integration.

So you can fix some $a$ and divide your integral into $0$ to $a$ and $a$ to $\sqrt{n}$ portions. The second portion will be bounded by $\int_a^{\infty}e^{-x^2}\,dx$ for all $n$ by the above. On the first portion, since $n\ln(1- {x^2 \over n}) = -x^2 + O({x^4 \over n})$, the function $n\ln(1- {x^2 \over n})$ converges uniformly to $-x^2$. Taking exponentials $(1 - {x^2 \over n})^n$ converges to $e^{-x^2}$ uniformly on $0 \leq x \leq a$. So you can apply the uniform convergence theorem on this first part. The second integral is never more than $\int_a^{\infty}e^{-x^2}\,dx$ by the above. So letting $a$ go to infinity gets you the result.

Zarrax
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$$I_n= \int_0 ^{\sqrt{n}} \left( 1-\frac{x^2}{n} \right)^ndx$$ let $x= \sqrt{n} t $ then $dx=\sqrt{n} dt$

$$I_n= \sqrt{n} = \int_0 ^{1} \left( 1-t^2 \right)^ndt$$ let $t =\sin(u)$ then $dt = \cos(u)du$

$$I_n= \sqrt{n} \int_0 ^{\frac{\pi}{2}} \cos^{2n+1}(u)du$$

by Wallis product $I_n =\sqrt{n} \frac{(2n)!!}{(2n+1)!!}$ by Stirling's approximation $$\lim_{n \to \infty} I_n=\lim_{n \to \infty} \sqrt{n} \frac{\sqrt{2\pi n}}{\sqrt{4n+2}} \left(\frac{2n}{e}\right)^n \left(\frac{e}{2n+1}\right)^{n+\frac{1}{2}}=\lim_{n \to \infty}\frac{\sqrt{n}}{\sqrt{2n+1}} \frac{\sqrt{2\pi n}}{\sqrt{4n+2}} \sqrt{e} \left( 1-\frac{1}{2n+1}\right)^n $$ $$\lim_{n \to \infty} = \frac{\sqrt{\pi}}{2} $$

$$\int_{0}^{\infty} e^{-x^2}dx= \frac{1}{2} \int_{-\infty}^{\infty} e^{-x^2}dx= \frac{1}{2}\sqrt{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy}=\frac{1}{2}\sqrt{\int_{0}^{2\pi} \int_{0}^{\infty} re^{-r^2}drd\theta}=\frac{\sqrt{\pi}}{2} $$

this Proves $\lim\limits_{n \to \infty} \int_{0}^{\sqrt n}(1-\frac{x^2}{n})^ndx=\int_{0}^{\infty} e^{-x^2}dx =\frac{\sqrt{\pi}}{2} $

pie
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