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Consider this integral: $$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz$$ How would you compute it?


I already solved this problem this way: $$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz = \left( \int_\infty^\infty e^{-x^2} \right)^3 = \pi^{3/2}$$

But I wanted to find it using substitution (spherical coordinates) but this is all I could do: $$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz = \lim_{j\,\rightarrow\infty}\int_0^jdu\int_0^\pi dv\int_0^{2\pi} e^{-u^2}u^2\sin(v) dw=$$

$$=2\pi\lim_{j\,\rightarrow\infty}\int_0^jdu\int_0^\pi e^{-u^2}u^2\sin(v)dv=4\pi\lim_{j\,\rightarrow\infty}\int_0^je^{-u^2}u^2du$$

But it doesn't get me anywhere. Help would be very appreciated, thanks in advance.

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  • $\begingroup$ That should be $e^{-u^2}$. The integrals over $v$ and $w$ produce a factor of $4 \pi$. $\endgroup$ – Ron Gordon May 15 '13 at 20:24
  • $\begingroup$ Yes, I changed the $e^{u^2}$ into $e^{-u^2}$, but the factor $4\pi$ was there from the beginning unless you I missed something. $\endgroup$ – user41489 May 15 '13 at 20:31
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To finish off your problem, you only need

$$\int_{0}^{\infty}e^{-x^2}x^2dx=\frac{\sqrt{\pi}}{4}$$

which can be shown by integrating

$$\int_{0}^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$$

by parts. That is, we have:

$$\frac{\sqrt{\pi}}{2}=\int_{0}^{\infty}e^{-x^2}dx=xe^{-x^2}\vert_{0}^{\infty}+2\int_{0}^{\infty}e^{-x^2}x^2dx=2\int_{0}^{\infty}e^{-x^2}x^2dx$$

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  • $\begingroup$ Very clear, thank you. $\endgroup$ – user41489 May 15 '13 at 20:42
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$$\int_0^{\infty}e^{-x^2}x^2dx = \frac 1 2 \int_0^{\infty } e^{-x^2} x \; 2x dx = \frac 1 2 \int_0^{\infty } e^{-u} \sqrt u du = \frac{\Gamma (3/2)}{2} = \frac 1 4\sqrt{\pi}$$

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