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

I need some help calculating the above limit.

What i have observed so far is that:

  1. For all x the limit of the sequence inside the integral as n tends to infinity is $e^{-x^2}$
  2. I can use Dini's theorem to show that: $f_n(x) = {(1-\frac{x^2}{n})^n}$ uniformly converges to $e^{-x^2}$
  3. Consequently i can use the theorem regarding "the integral of the limit is the limit of integrals" for the function sequence $f_n(x)$ and it's limit function $f(x) = e^{-x^2}$

All this is well and good, but i would be ignoring the fact that the interval integrated upon is [0,$\sqrt{n}$], and so also affected by the limit.

What should i do to resolve this?

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marked as duplicate by Davide Giraudo, Byron Schmuland, Sasha, J. M., Jason DeVito Aug 7 '12 at 12:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers 2

up vote 2 down vote accepted

We can apply dominated convergence theorem, since $e^{—x}\geq 1-x$ for each $x\geq 0$ and $e^{-x^2}$ is integrable on $(0,+\infty)$. Define $g(x)=e^{-x}-(1-x)$. Then $g'(x)=-e^{-x}+1\geq 0$ for $x\geq 0$ hence $g(x)\geq g(0)=0$. Now, we have for $0\leq x\leq \sqrt n$, $$0\leq \left(1-\frac{x^2}n\right)^n\leq \left(e^{-\frac{x^2}n}\right)^n=e^{-x^2}.$$

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Could you please elaborate on your answer? I am not sure what you mean. –  JohnnyE. Aug 6 '12 at 20:22
    
@JohnnyE. I have added more details. –  Davide Giraudo Aug 6 '12 at 20:26
1  
Thanks. also apparently this question has been asked before: math.stackexchange.com/questions/91378/… –  JohnnyE. Aug 6 '12 at 20:29
    
I am still not clear how to use what you wrote to show that: $\lim\limits_{n \to \infty} \int_{0}^{\sqrt n}(1-\frac{x^2}{n})^ndx=\int_{0}^{\infty} e^{-x^2}dx$ , do you mean to use the sandwich theorem? if so, what do i do about the lower bound? –  JohnnyE. Aug 6 '12 at 21:35
    
No, I meant the dominated convergence theorem (I show that the function we integrate are bounded independently on $n$ by an integrable function). –  Davide Giraudo Aug 6 '12 at 21:40

A related problem. Apply the substitution of variables $ y = x^2 $ to your integral gives

$$ \frac{1}{2}\,\int _{0}^{n}\! \left( {\frac {n-y}{n}} \right) ^{n}{\frac {1}{ \sqrt {y}}}{dy} $$

Apply another change of variables $ y=nz $ casts your integral to the beta function

$$ \frac{1}{2}\,\sqrt {n}\int _{0}^{1}\!{\frac { \left( 1-z \right) ^{n}}{\sqrt { z}}}{dz} = \frac{1}{2}\,{\frac {\sqrt {n}\Gamma \left( n+1 \right) \Gamma(\frac{1}{2})}{\Gamma \left( n+3/2 \right) }} $$

Taking the limit as $n\rightarrow \infty$ gives $ \frac{1}{2} \Gamma(\frac{1}{2}) = \frac{\pi}{2}\,.$

You can use Stirling's approximation $n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$ of the $n!=\Gamma(n+1)$ to evaluate the above limit.

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