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The question is to prove: $$I=\int_{-\infty}^{\infty}e^{-x^2}\,\Bbb dx=\sqrt{\pi}.$$

Let \begin{align} I_r&=\int_{-r}^{r}e^{-x^2}\,\Bbb dx \\ \implies I_r^2&=\int_{-r}^{r}e^{-x^2}\,\Bbb dx\int_{-r}^{r}e^{-y^2}\,\Bbb dy \\ &=\iint_{[-r,r]^2} e^{-x^2-y^2}\,\Bbb dx\,\Bbb dy. \end{align}

$x^2+y^2\leq r^2=D_r$

$$K_r=\iint_{D_r}{e^{-x^2-y^2}}\,\Bbb dx\,\Bbb dy$$

$$0\leq I_r-K_r=\int_{Q_r \setminus D_r}e^{-x^2-y^2}\,\Bbb dx\,\Bbb dy\leq e^{-r^2}\mu(Q_r)=e^{-r^2}4r^2\rightarrow0; r\rightarrow \infty.$$ The rest is clear to me just this last inequality: $\leq e^{-r^2}\mu(Q_r)$, where $\mu(A)$ is a measure of $A$. I’ll type out the rest of the proof if need be, for educational purposes$\ldots$

Rócherz
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Bozo Vulicevic
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2 Answers2

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I'm not sure if you are getting at the usual proof of this, which is $$ \begin{align} \left(\int_{-R}^R e^{-x^2}\,\mathrm{d}x\right)^2 &=\int_{-R}^R e^{-x^2}\,\mathrm{d}x\int_{-R}^R e^{-y^2}\,\mathrm{d}y\\ &=\int_{-R}^R\int_{-R}^R e^{-x^2-y^2}\,\mathrm{d}x\,\mathrm{d}y \end{align} $$ and by sandwiching the square between two circles, $$ \int_0^{2\pi}\int_0^Re^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta\le\int_{-R}^R\int_{-R}^R e^{-x^2-y^2}\,\mathrm{d}x\,\mathrm{d}y\le\int_0^{2\pi}\int_0^{\sqrt2R}e^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta $$ which combined with $$ \int_0^{2\pi}\int_0^Re^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta=\pi\left(1-e^{-R^2}\right) $$ and the Squeeze Theorem yields $$ \left(\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x\right)^2=\pi $$

robjohn
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  • Why use the squeeze theorem when one can just use polar coordinates to directly find the result? – Ant Apr 10 '15 at 23:08
  • @Ant: The whole point is to show that the limit of the integral over the square (given by the square of the integral) is equal to the limit of the integral over the circle (given by the polar integral). This is done by sandwiching the square of side $2R$ between the circles of radius $R$ and $\sqrt2R$. – robjohn Apr 10 '15 at 23:12
  • I'm aosrry but I don't understand.. Once you get to $\int_0^{2\pi}\int_0^Re^{-r^2}r,\mathrm{d}r,\mathrm{d}\theta$ can't you just solve it like $= 2\pi \left|- e^{-r^2}/2\right| ^R_0 = \pi (-e^{-R^2}+1) \to \pi$ ? – Ant Apr 11 '15 at 10:46
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    @Ant: how do you know that the integral on expanding disks is equal to the integral on expanding squares? There are many examples of integrals that converge to different values depending on how the domain of integration expands (however, most, if not all, of those examples involve cancellation between positive and negative values). The idea of the argument in the question is to sandwich the integral so that we know that the limit on squares equals the limit on disks. – robjohn Apr 11 '15 at 13:39
  • @Ant: Another approach would be to use the Lebesgue's Monotone Convergence Theorem to show that no matter how we grow the domain, we reach the same value. I assume that the OP does not know this theorem. – robjohn Apr 11 '15 at 13:40
  • Ah I see. thank you :-) – Ant Apr 11 '15 at 13:44
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I would go for some integral mean value theorem: $$ \int\limits_{Q_r \setminus D_r} e^{-x^2 -y^2} \,dx \,dy = e^{-(X^2+Y^2)} \int\limits_{Q_r \setminus D_r} \,dx \,dy = e^{-(X^2+Y^2)} \mu (Q_r \setminus D_r) \le e^{-(X^2 + Y^2)} \mu(Q_r) $$ where $(X,Y) \in Q_r \setminus D_r$ ("square with circular hole") and the minimum of $X^2+Y^2$ is at $\partial D_r$ with $r^2$, so $$ e^{-(X^2 + Y^2)} \le e^{-r^2} $$ for all $(X,Y) \in Q_r \setminus D_r$. This leads then to $$ e^{-(X^2 + Y^2)} \mu(Q_r) \le e^{-r^2} \mu(Q_r) $$

mvw
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