n-dimensional Fourier integral let be $ \int_{R^{n}}dVf(r)e^{i(k.r)} $ the n-dimensional Fourier integral.
$ dV=dxdydz.... $ the volume and $ (k.r)= \sum_{n} k_{n}.x_{n} $ is the scalar product of the position vector 'r' and the vector 'k'
if the function $ f(r) $ depends only on the radial coordinate (but not on the angles) how can i reduce the n-dimensional fourier integral to the evaluation of the one dimensional integral
$ \int_{0}^{\infty}drf(r)r^{n-1}H(kr) $ , what is the definition of $ H(kr) $ ??
thanks.
 A: The derivation is similar to the one found at Hankel transform.
Let $r=||\mathbf{r}||$, $k=||\mathbf{k}||$ and
$F(\mathbf{k}):=\iint f(r)
e^{i\mathbf{k}\cdot\mathbf{r}}\,d\mathbf{r}$.
With no loss of generality, we can pick a hyperspherical coordinate system $(r, \theta,\ldots)$ such that the $\mathbf{k}$ vector lies on the $\theta = 0$ axis. The Fourier transform is now written in these hyperspherical coordinates as
$$F(\mathbf{k})=\int_{r=0}^\infty
\int_{\theta=0}^{\pi}f(r)e^{ikr\cos \theta }\,r^{n-1}S_{n-2}(\sin \theta)\,dr\,d\theta$$
Using $S_{n-2}(\sin \theta)=S_{n-2}(1)\sin^{n-2} \theta$ and changing integration order, we get
$$F(\mathbf{k})=\int_{r=0}^\infty
f(r)r^{n-1}(S_{n-2}(1)\int_{\theta=0}^{\pi}e^{ikr\cos \theta }\,\sin^{n-2} \theta\,d\theta)\,dr$$.
So $H(kr)=S_{n-2}(1)\int_{\theta=0}^{\pi}e^{ikr\cos \theta }\,\sin^{n-2} \theta\,d\theta$. Note that it would be better to write $H_n(kr)$ instead of $H(kr)$. For $n=2$, we have $S_0(1)=2$ and $\int_{\theta=0}^{\pi}e^{ikr\cos \theta }\,d\theta=\pi J_0(kr)$, so we get back the well known Fourier-Bessel transform.
