How to calculate following integral:

$$\lim_{n\rightarrow\infty}\int_{-\sqrt{n}}^{\sqrt{n}}{\left(1-\frac{x^2}{2n}\right)^n}dx$$ Prove that this integral exists and compute its value.

I just do not how start it. It is in real analysis, but I did not see any relation with real analsis.

  • 5
    $\begingroup$ Why don't you just evaluate the integral for fixed $n$ and then compute the limit? $\endgroup$ – Guest Apr 4 '15 at 22:30
  • $\begingroup$ And of course, it has a relation with real analysis...!! $\endgroup$ – Daniel Apr 4 '15 at 22:31
  • $\begingroup$ I guess the question becomes easier if one labels it "freshman calculus" instead of "real analysis". $\endgroup$ – Andreas Blass Apr 4 '15 at 22:34
  • $\begingroup$ @Guest et al., the OP has edited the integrand. It's no longer quite so simple. $\endgroup$ – Barry Cipra Apr 4 '15 at 22:49
  • 1
    $\begingroup$ See Gaussian integral. $\endgroup$ – Lucian Apr 4 '15 at 22:57

The integrals may be written $\int_{-\infty}^\infty f_n(x)\, dx$, where $f_n(x) = 1_{[-\sqrt{n},\sqrt{n}]}(x)\cdot(1 - \frac{x^2}{2n})^n$. Now since $1 - \frac{x^2}{2n} \le e^{-x^2/2n}$, $|f_n(x)| \le e^{-x^2/2}$. Furthermore, $\lim_{n\to \infty} f_n(x) = e^{-x^2/2}$ and $\int_{-\infty}^\infty e^{-x^2/2}\, dx$ is finite. By the dominated convergence theorem, your limit is $\int_{-\infty}^\infty e^{-x^2/2}\, dx = \sqrt{2\pi}$.


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