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I'm preparing for a calculus exam, I'd like help in solving this question.

Let $x \in \mathbb R^n$, $|x|={(x_1^2+x_2^2+...+x_n^2)^{\frac{1}{n}}}$,

Show that $$\int_{\mathbb R^n} e^{|x|^{-n}}dx$$ is equal to the volume of the unit sphere of $\mathbb R^n$.

The way to do this in my opinion is write the integral as

$$\int_{\mathbb R^n}e^{\frac{1}{x_1^2+x_2^2+...+x_n^2}}dx$$

And then go to spherical coordinate as shown here http://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates

But I haven't managed to get a definite result.

Would someone please assist?

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Hint: if $x=r\omega$, where $r>0$ and $|\omega|=1$ then $dx=r^{N-1}drd\omega$ and $$\int_{\mathbb{R}^N}e^{-|x|^n}dx=\int_{S_1}\int_0^\infty e^{-r^N}r^{N-1}drd\omega,$$

where $S_1$ is the unit sphere.

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  • $\begingroup$ There seems to be a problem. The $dr$ integral is equal to $\frac{1}{n}$ and the $dw$ integral is equal to $V(S_1)$. so we get that the initial integral we wanted to show is equal to $V(S_1)$, is actually equal to $\frac{1}{n}V(S_1)$ $\endgroup$ – Oria Gruber Jun 11 '14 at 22:06
  • $\begingroup$ I will give you some time to think. I will only point out now that my answer is not wrong. In fact, your comment is not quite right. $\endgroup$ – Tomás Jun 11 '14 at 22:17
  • $\begingroup$ The $\int_{0}^{\infty}e^{-r^n}r^{n-1}dr=\frac{1}{n}$ of this I am sure. I checked it with a computer on several $n$s and proved it on paper. If my first comment is wrong, then $\int_{S_1}1dw \neq V(S_1)$. $\endgroup$ – Oria Gruber Jun 11 '14 at 22:23
  • $\begingroup$ You are confusing surface area with volume. $\endgroup$ – Tomás Jun 11 '14 at 22:24
  • $\begingroup$ Then clearly I misunderstood something. I thought that by definition $\int_A1dv$ is the volume of $A$. much like how $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}1dxdydz$ is the volume of the unit cube. $\endgroup$ – Oria Gruber Jun 11 '14 at 22:35

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