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We should express the integral $I_{n}=\int_{\mathbb{R}^{n}}\exp\left(\frac{-\left\Vert x\right\Vert ^{2}}{2}\right)\mathrm{d}x$ using $I_1$. Where $\left\Vert x\right\Vert =\left(x_{1}^{2}+\cdots x_{n}^{2}\right)^{\frac{1}{2}}$ is the Euclidean norm.

We have this excursion to somewhat "advanced analysis" in otherwise "discreet" subject. My analysis skills are wery weak and I would much apperciate any help.

My progress: (probably wrong, see bellow)

$\int_{\mathbb{R}^{n}}\exp\left(\frac{-\left\Vert x\right\Vert ^{2}}{2}\right)\mathrm{d}x=\int_{\mathbb{R}^{n}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n}^{2}\right)}{2}\right)\mathrm{d}x=\int_{\mathbb{R}^{n}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}+\frac{-x_{n}^{2}}{2}\right)\mathrm{d}x=\int_{\mathbb{R}^{n}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}\right)\cdot\exp\left(\frac{-x_{n}^{2}}{2}\right)\mathrm{d}x$

This would lead me to integration per partes using $u'=\exp\left(\frac{-x_{1}^{2}}{2}\right)$ and $v=\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}\right)$ because then $u=\int \exp\left(\frac{-x_{1}^{2}}{2}\right) \mathrm{d}x_1$ looks somewhat similar to $I_{1}=\int_{\mathbb{R}}\exp\left(\frac{-x^{2}}{2}\right)\mathrm{d}x$ which is available.

I am not however able to formally complete the process. I can only compute single variable indefinite integrals.

Thank you very much.


As was suggested in an answer, a possible solution might be:

$$\begin{align*} \int_{\mathbb{R}^{n}}\exp\left(\frac{-\left\Vert x\right\Vert ^{2}}{2}\right)\mathrm{d}x &= \int_{\mathbb{R}^{n}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n}^{2}\right)}{2}\right)\mathrm{d}x=\int_{\mathbb{R}^{n}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}+\frac{-x_{n}^{2}}{2}\right)\mathrm{d}x \\&=\int_{\mathbb{R}^{n}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}\right)\cdot\exp\left(\frac{-x_{n}^{2}}{2}\right)\mathrm{d}x \\&=\int_{\mathbb{R}^{n-1}}\int_{\mathbb{R}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}\right)\cdot\exp\left(\frac{-x_{n}^{2}}{2}\right)\mathrm{d}x_{n}\mathrm{d}x_{n-1}\ldots\mathrm{d}x_{1} \\&\overset{Fubini}{=}\int_{\mathbb{R}^{n-1}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}\right)\left(\int_{\mathbb{R}}\exp\left(\frac{-x_{n}^{2}}{2}\right)\mathrm{d}x_{n}\right)\mathrm{d}x_{n-1}\ldots\mathrm{d}x_{1} \\&=\int_{\mathbb{R}^{n-1}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}\right)I_{1}\mathrm{d}x_{n-1}\ldots\mathrm{d}x_{1} \\&=I_{1}\int_{\mathbb{R}^{n-1}}\exp\left(\frac{-\left(x_{1}^{2}+\cdots x_{n-1}^{2}\right)}{2}\right)\mathrm{d}x_{n-1}\ldots\mathrm{d}x_{1} \\&=I_{1}I_{n-1} \end{align*}$$

However, I am still not sure whether this is a formally correct solution.

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  • $\begingroup$ You are nearly done. $I_1$ is an integral - you are not asked to evaluate it, just write $I_n$ in terms of it. Note that the dx in $I_n$ means $dx_{1}dx_{2}...dx_{n}$. $\endgroup$ – Paul Apr 4 '14 at 13:08
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Here $x=(x_1,\dots,x_n)$ and $\mathrm dx$ actually means $\mathrm dx_1\dots\mathrm dx_n$.

Using Fubini-Tonelli's theorem, we find $I_n=I_{n-1}I_1$, hence $I_n=I_1^n$.

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  • $\begingroup$ Thank you! It took me a while to absorb. Would you mind checking my solution using Fibini's theorem? $\endgroup$ – discrete Apr 4 '14 at 13:59

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