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Say I have a function $p_v(r) \in L^2(\mathbb{R})$ given by $$p_v(r) = \int_0^\infty P(k) J_v(rk)\,k\,dk$$

From mucking around in MATLAB it seems the following is true: $$\int_{r=0}^\infty \int_{\theta=0}^{\pi/n} e^{i n \theta} p_n(r)\,r\,dr\,d\theta = \int_{r=0}^\infty \int_{\theta=0}^{\pi/m} e^{i m \theta} p_m(r)\,r\,dr\,d\theta$$

Which would mean $$\frac{1}{n}\int_{r=0}^\infty p_n(r)\,r\,dr = \frac{1}{m}\int_{r=0}^\infty p_m(r)\,r\,dr$$

Assuming I haven't done something wrong, how do I show this is true? Tried using the integral form of the bessel function and other random relationships came across but I'm stumped. Furthermore, is there an expression for $p_v(r)$ when $P(k) = 1$?

edit: Fix thanks to @oenamen


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up vote 2 down vote accepted

The equation you found, \begin{equation} \int_{r=0}^\infty \int_{\theta=0}^{\pi/n} e^{i n \theta} p_n(r)\,r\,dr\,d\theta = \int_{r=0}^\infty \int_{\theta=0}^{\pi/m} e^{i m \theta} p_m(r)\,r\,dr\,d\theta,\tag{1} \end{equation} (which happens to be true in a restricted sense, see below) does not imply $\int_{0}^\infty p_n(r)\,r\,dr = \int_{0}^\infty p_m(r)\,r\,dr$. Just do the angular integrals. You will find instead that it implies $$\frac{1}{n} \int_{0}^\infty p_n(r)\,r\,dr = \frac{1}{m}\int_{0}^\infty p_m(r)\,r\,dr.$$ We can check this explicitly, $$\begin{eqnarray} \frac{1}{n} \int_{0}^\infty p_n(r)\,r\,dr &=& \frac{1}{n} \int_0^\infty P(k)\,k \,dk \int_0^\infty J_n(k r)\,r\,dr \\ &=& \frac{1}{n} \int_0^\infty P(k)\,k \,dk \frac{n}{k^2} \\ &=& \int_0^\infty P(k)\,k^{-1} \,dk. \end{eqnarray}$$ The final integral is just some constant, independent of $n$. $P(k)$ must of course be well enough behaved so the integral makes sense.

Here we have used the fact that the Hankel transform of $r^s$ is $$\frac{\Gamma\left(1+\frac{1}{2}(n+s)\right)}{\Gamma\left(\frac{1}{2}(n-s)\right)} \frac{2^{s+1}}{k^{s+2}}.$$ Thus, for example, the Hankel transform of $1$ (used above) is $n/k^2$.

Addendum: Notice that, strictly speaking, the integral $$\int_0^\infty r^{s+1} J_n(k x) dr = \frac{\Gamma\left(1+\frac{1}{2}(n+s)\right)}{\Gamma\left(\frac{1}{2}(n-s)\right)} \frac{2^{s+1}}{k^{s+2}}$$ is divergent unless $-\mathrm{Re}\, n-2< \mathrm{s} <-\frac{1}{2}$ and $k>0$. These conditions are not satisfied for $s=0$. However, we can define the value of the integral to be the analytic continuation of the above formula to $s=0$. In fact, this practice is quite common and useful in physics. Thus, strictly speaking each side of (1) is divergent, but in a sense it is still true!

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Thanks heaps. The nth order Hankel transform of r^s was what I was missing (and the 1/n part ..fp) - do you have a reference for that? I have looked all through Gradshteyn and Ryzhic and couldnt find it. – geometrikal Apr 8 '12 at 0:17
@geometrikal: Glad to help. The transform can probably be derived using one of the integral representations of the Bessel function. You can also find it in the "Combinations of Bessel functions and powers" section of Gradshteyn and Ryzhik (6th). It is equation 6.561-14. – user26872 Apr 8 '12 at 0:29
Looking at 6.561-14 (with v=n x=k mu=s) has [-Re n - 1 < s < 1/2, k > 0]. Since s = 1 does it apply? Thanks. – geometrikal Apr 8 '12 at 1:16
Thanks again, this is beyond my current abilities, but I think I can visualise whats going on - they are diverging at the same 'speed' w.r.t $r$(?). I'm guessing that the reason I got a non-infinite number when mucking around in MATLAB is due to bounding. Will try to reformulate it that way. – geometrikal Apr 8 '12 at 4:23
@geometrikal: It is a good exercise to check that if the analytically continued formula for the transform of $r^s$ is used as the definition of the Hankel transform of $r^s$, (and a similar formula is used for the definition of the inverse Hankel transform, with $k$ replaced by $r$) the inverse transform back to $r$-space gives you $r^s$, as it should! – user26872 Apr 8 '12 at 5:24

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