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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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$H$ should probably be some Hankel transform. –  Mark Dec 30 '11 at 9:56
    
There are generalized polar coordinates where you view $\mathbb R^n\setminus \{0\} \cong \mathbb R^+ \times S^{n-1}$, sending a vector $v$ to $(||v,||,v/||v||)$. Now, use Fubini's theorem. –  Aaron Dec 30 '11 at 9:56
    
Did you try to use hyperspherical coordinates: en.wikipedia.org/wiki/Hypersphere. –  Davide Giraudo Dec 30 '11 at 9:56
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however the quantity $ (k.r) $ after the change to polar coordinates should depend on an angle i think that $ (k.r) = |k|.|r|cos(\theta) $ where 'theta' is the angle (in polar coordinates) between the 2 vectors k and r . –  Jose Garcia Dec 30 '11 at 10:02
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Yes, but the integral of $f(z,r)=\int_{S^{n-1}}e^{i(z\cdot ru)}du$ depends only on $r$ and the length of $z$ (as can be seen by symmetry). The only question is how it scales as $r$ and the length of $z$ change. –  Aaron Dec 30 '11 at 10:07
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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.

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