# How to show integral of different order Hankel transformed functions are equal?

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

Thanks.

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The equation you found, $$\int_{r=0}^\infty \int_{\theta=0}^{\pi/n} e^{i n \theta} p_n(r)\,r\,dr\,d\theta = \int_{r=0}^\infty \int_{\theta=0}^{\pi/m} e^{i m \theta} p_m(r)\,r\,dr\,d\theta,\tag{1}$$ (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!
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