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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.

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    Do you need to have an explicit, non-integral form of $H$, or is it enough to express $H$ as an integral? If you wanted to evaluate the integral explicitly, you could do so as an iterated integral in hyperspherical coordinates using lots of sins and cosines. Once you are in coordinates, you can probably find an inductive formula depending on $n$. If you don't need a form for $H$ more explicit than an integral, there is enough information in the comments to construct a reasonable outline of a solution (minus change of variables).2011-12-30

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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.