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.
 A: 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! 
