I want to prove that: \begin{equation} \lim_{t\to 0} \int_{|x|>\epsilon} \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2 \pi t}} dx=0, \end{equation} for any $\epsilon >0$

I've shown using polar coordinates that \begin{equation} \int \limits_{-\infty}^{+\infty} \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2 \pi t}}=1. \end{equation}

I tried the same approach for the limit I need to prove. But I obtained a result that depended on $\epsilon$ only, and not on $t$, and therefore it does not go to $0$. Where is my mistake?

Other ways to approach it would be greatly appreciated too (preferably, with real analysis methods only).

  • $\begingroup$ it is very difficult to answer what goes wrong in your computation that gave you "a result that depended on $\epsilon$ only" unless you actually show that computation. $\endgroup$ – hmakholm left over Monica May 5 '15 at 23:38
  • $\begingroup$ The computation I did is similar with the one here: mathworld.wolfram.com/GaussianIntegral.html $\endgroup$ – odnerpmocon May 5 '15 at 23:41
  • $\begingroup$ The only important difference is that $r$ is from $2 \epsilon ^2$ to $\infty$ instead of from 0 to $\infty$ $\endgroup$ – odnerpmocon May 5 '15 at 23:42

We can use the same polar coordinates trick here and in fact obtain a good bound on the integral. Let $$I(\epsilon,t) = \int_{\vert x \vert > \epsilon} \dfrac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}dx$$ We now have $$I^2(\epsilon,t) = \int_{\vert x \vert > \epsilon} \dfrac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}dx \cdot \int_{\vert y \vert > \epsilon} \dfrac{e^{-\frac{y^2}{2t}}}{\sqrt{2\pi t}}dy = \int_{\vert x \vert > \epsilon} \int_{\vert y \vert > \epsilon} \dfrac{e^{-\frac{x^2+y^2}{2t}}}{2\pi t}dxdy \leq \int_{x^2+y^2 > \epsilon^2} \dfrac{e^{-\frac{x^2+y^2}{2t}}}{2\pi t}dxdy$$ Now moving to polar coordinates, we have $$I^2(\epsilon,t) \leq \int_{\theta=0}^{2\pi} \int_{r=\epsilon}^{\infty} \dfrac{e^{-\frac{r^2}{2t}}}{2\pi t}rdrd\theta = \dfrac1t \int_{r=\epsilon}^{\infty}re^{-\frac{r^2}{2t}}dr = \dfrac1t \cdot te^{-\frac{\epsilon^2}{2t}} = e^{-\frac{\epsilon^2}{2t}}$$ This gives us that $$\boxed{\color{blue}{I(\epsilon,t) \leq e^{-\frac{\epsilon^2}{4t}}}}$$ Take $\lim t \to 0$ to obtain what you want.

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  • $\begingroup$ Thanks. My approach was pretty much similar, other than a change of variable in the beginning which I shouldn't have done. $\endgroup$ – odnerpmocon May 5 '15 at 23:49

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