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 Post subject: Multiple integral
PostPosted: Mon Jun 19, 2017 8:59 pm 
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Let $\langle \cdot, \cdot \rangle$ denote the usual inner product of $\mathbb{R}^m$. Evaluate the integral

$$\mathcal{M} = \int \limits_{\mathbb{R}^m} \exp \left( - ( \langle x, \mathcal{S} ^{-1} x \rangle )^a \right) \, {\rm d}x$$

where $\mathcal{S}$ is a positive symmetric $m \times m$ matrix and $a>0$.

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 Post subject: Re: Multiple integral
PostPosted: Thu Nov 30, 2017 10:00 pm 

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Since $\mathcal{S}$ is a positive symmetric matrix , so is $\mathcal{S}^{-1}$. For a positive symmetric matrix $\mathcal{A}$ there exists an $\mathcal{R}$ positive symmetric matrix such that $\mathcal{A} = \mathcal{R}^2$. Applying this to $\mathcal{S}^{-1}$ our integral becomes

\[\mathcal{M} = \int \limits_{\mathbb{R}^m} \exp \left ( - \left \| \mathcal{R} x \right \|^{2a} \right ) \, {\rm d}x\]

where $\left \| \cdot \right \|$ is the Euclidean norm. Applying a change of variables we have that

\[\mathcal{M} = \det \left ( \mathcal{R}^{-1} \right ) \int \limits_{\mathbb{R}^m} e^{-\left \| y \right \|^{2a}} \, {\rm d}y\]

Since $\det \left ( \mathcal{R}^{-1} \right ) = \sqrt{\det \left ( \mathcal{S} \right )}$ then by converting to polar coordinates we have that

\begin{align*} \mathcal{M} &= \omega_m \sqrt{\det \left ( \mathcal{S} \right )} \int_{0}^{\infty} r^{m-1} e^{-r^{2a}} \, {\rm d}r \\ &= \frac{\omega_m}{m} \sqrt{\det \left ( \mathcal{S} \right )}\Gamma \left ( \frac{m}{2a} + 1 \right ) \end{align*}

Here $\omega_m$ denotes the surface area measure of the unit sphere and it is known to be

\[\omega_m = \frac{2 \pi^{m/2}}{\Gamma \left ( \frac{m}{2} \right )}\]

hence

\[\mathcal{M}=\frac{\sqrt{\det (\mathcal{S}) }\pi^{m/2}\Gamma\left(\frac{m}{2a}\right)}{2^{a-1}\Gamma\left(\frac{m}{2}\right)}\]

where $\Gamma$ denotes the Gamma Euler function for which it holds that

\[\Gamma(x+1) = x \Gamma(x) \quad \text{forall} \quad x>0\]

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