# Polar coordinates in $d$ dimensions

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