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I'm having troubles with the following integral: $$ \int_{|\vec{\sigma}|^2=\epsilon^2}d\vec{\sigma}\,\exp\left({a \,\vec{m}\cdot \vec{\sigma}}\right)\,, $$ where $\vec{\sigma},\,\vec{m}\in\mathbb{R}^d$, $a,\,\epsilon$ are positive constants. I'm looking for a general solution (in terms of Modified Bessel functions) in $d$ dimensions.

I started with $d=2$: $$ \int_{\sigma_x^2+\sigma_y^2=\epsilon^2} d\sigma_x d\sigma_y \exp{\left(a (m_x\sigma_x + m_y\sigma_y)\right)}\stackrel{?}{=}\epsilon\int_0^{2\pi}d\theta \exp{\left( a |m|\epsilon \cos(\theta)\right)}=2\pi\epsilon\,\mathcal{I}_0(a |m|\epsilon) \, $$ where $\mathcal{I}_0(x)$ is the Modified Bessel function of the first kind. Is the $d=2$ case right or I made some mistakes? I used the polar coordinates in the first integral (from $d\sigma_xd\sigma_y$ to $d\theta$) writing $\vec{m}\cdot\vec{\sigma}=|m|\epsilon \cos({\theta_m-\theta})$, indicating with $\theta_m$ the angle of the vector $\vec{m}$ in $\mathbb{R}^2$. Then I used the fact the integrand is periodic and the support of the integral is exactly equal to the period, so the integral doesn't depend on $\theta_m$, getting the result.

In generic dimension $d$ I can use hyperspherical coordinates: $$ \int_{|\vec{\sigma}|^2=\epsilon^2}d\vec{\sigma}\,\exp\left({a \,\vec{m}\cdot \vec{\sigma}}\right)= \epsilon^{d-1}\int_0^\pi \prod_{i=1}^{d-2}d\phi_{i} (\sin{\phi_i})^{d-1-i}\int_0^{2\pi}d\phi_{d-1} \exp{(a|m|\epsilon \cos{(?)})}\, $$ is the argument of the $\cos$ $\phi_{d-1}$ or $\phi_1$? I think it is $phi_1$ but in this case I cannot understand what I did for the $d=2$ case, I am getting confused.

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1 Answer 1

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I'll provide the following answer.

  1. The result of the $d=2$ case is right, however there are some problems in the derivations. In particular, when passing to the polar coordinates, introducing the $\cos$ of the angle between $\theta_m$ and $\theta$, the integral becomes $2\int_0^{\pi} d\theta \exp{(a |m| \epsilon \cos{(\theta)})}$, since the difference of two angles is always between $0$ and $\pi$ radians. Then the procedure is the same as I showed.

  2. The general case can be treated as follows.

$$ \int_{|\vec{\sigma}|^2=\epsilon^2}d\vec{\sigma}\,\exp\left({a \,\vec{m}\cdot \vec{\sigma}}\right)= \epsilon^{d-1}\int_0^\pi \prod_{i=1}^{d-2}d\phi_{i} (\sin{\phi_i})^{d-1-i}\int_0^{2\pi}d\phi_{d-1} \exp{(a|m|\epsilon \cos{(\phi_1)})}\, $$ which leads to $$ \mathcal{S}_{d-2}\,\epsilon^{d-1}\int_0^\pi d\phi_{1} (\sin{\phi_1})^{d-2} \exp{(a|m|\epsilon \cos{(\phi_1)})}\, $$ where $\mathcal{S}_{d-1}=\frac{2 \pi^{\frac{d}{2}}} {\Gamma\left(\frac{d}{2}\right)}$.

I am now wondering if there is a way to cast the last integral in the Modified Bessel function of the first kind.

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