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.


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .