Hankel transform of a Bessel function of different order Here I found that
$$
\int_0^\infty J_\nu(kr) J_\nu(sr) r dr = \frac{\delta(k - s)}{s} = \frac{1}{s^2}\delta\left(1 - \frac{k}{s}\right).
$$
I wonder how can that be derived and if a similar method can be applied to compute the Hankel transform of order $\mu$ of the $J_\nu(kr)$ function, that is
$$
\mathscr{H}_\mu[J_\nu(kr)](s) \equiv \int_0^\infty J_\nu(kr) J_\mu(sr) r dr = \frac{1}{s^2}\int_0^\infty J_\nu\left(\frac{k}{s} u\right) J_\mu(u) udu
$$
I found this integral in the H. Batemann's Tables of integral transforms (8.11.9):
$$
\int_0^\infty x^{-\lambda} J_\mu(ax) J_\nu(xy) dx = \\ =
\frac{\Gamma[(\mu+\nu-\lambda+1)/2]}
{2^\lambda}
\begin{cases}
\frac{y^\nu}{a^{\nu-\lambda+1} \Gamma[(\mu-\nu+\lambda+1)/2]}
{}_2\tilde{F}_1\left(
\frac{\mu+\nu-\lambda+1}{2},
\frac{\nu-\mu-\lambda+1}{2};\nu+1;\frac{y^2}{a^2}
\right), &0 < y < a\\
\frac{a^\mu}{y^{\mu-\lambda+1} \Gamma[(\nu-\mu+\lambda+1)/2]}
{}_2\tilde{F}_1\left(
\frac{\mu+\nu-\lambda+1}{2},
\frac{\mu-\nu-\lambda+1}{2};\mu+1;\frac{a^2}{y^2}
\right), &a < y < \infty\\
\end{cases}
$$
but it is for $\Re (\mu + \nu) + 1 > \Re \lambda > -1$ and I have exactly $\lambda = -1$. I suppose that is due to the fact that $\int_0^\infty J_\nu(kr) J_\mu(sr) r dr$ diverges in classic sense, but it may exist as a distribution, just like $\int_0^\infty J_\nu(kr) J_\nu(sr) r dr$ does.
 A: APPROACH 1:
We can show that $J_{\nu}(kr)$ and $J_{\nu}(sr)$ are orthogonal by appealing to the governing ODE 
$$\frac{d}{dr}\left(r\frac{dJ_{\nu}(kr)}{dr}\right)+\left(k^2r-\frac{\nu}{r}\right)J_{\nu}(kr)=0 \tag 1$$
$$\frac{d}{dr}\left(r\frac{dJ_{\nu}(sr)}{dr}\right)+\left(s^2r-\frac{\nu}{r}\right)J_{\nu}(sr)=0 \tag 2$$
Multiplying $(1)$ by $J_{\nu}(sr)$ and $(2)$ by $J_{\nu}(kr)$, subtracting, and integrating reveals
$$\begin{align}
(s^2-k^2)\int_0^{\infty}J_{\nu}(kr)J_{\nu}(sr)r\,dr&=\int_0^{\infty}\left(J_{\nu}(sr)\frac{d}{dr}\left(r\frac{dJ_{\nu}(kr)}{dr}\right)-J_{\nu}(kr)\frac{d}{dr}\left(r\frac{dJ_{\nu}(sr)}{dr}\right)\right)\,dr\\\\
&=\int_0^{\infty}\frac{d}{dr}\left(rJ_{\nu}(sr)\frac{dJ_{\nu}(kr)}{dr}-rJ_{\nu}(kr)\frac{dJ_{\nu}(sr)}{dr}\right)\,dr\\\\
&=0
\end{align}$$

APPROACH 2: For Integer Ordered $\nu$
We start with the two-dimensional Fourier_Transform pair
$$F(k_x,k_y)=\iint_{-\infty}^\infty f(x,y)e^{-ik_xx-ik_yy}\,dx\,dy \tag 3$$
$$f(x,y)=\frac{1}{(2\pi)^2}\iint_{-\infty}^\infty F(k_x,k_y)e^{ik_xx+ik_yy}\,dk_x\,dk_y \tag 4$$
Next, we convert the transform pair to cylindrical coordinates through the transformations $x=\rho \cos \phi$, $y=\rho \sin \phi$ and $k_x=k_{\rho}\cos \theta$, $k_y=k_{\rho}\sin \theta$.  Then, $(3)$ and $(4)$ become
$$\hat F(k_{\rho},\theta)=\int_0^\infty\int_0^{2\pi} \hat f(\rho,\phi)e^{-ik_{\rho}\rho\cos(\theta-\phi)}\,\rho\,d\rho\,d\phi \tag {3'}$$
$$\hat f(\rho,\phi)=\frac{1}{(2\pi)^2}\int_0^\infty\int_0^{2\pi} \hat F(k_{\rho},\theta)e^{ik_{\rho}\rho\cos(\theta-\phi)}\,k_{\rho}\,dk_{\rho}\,d\theta \tag {4'}$$
where we used the addition angle formuls $\cos \phi \cos \theta +\sin \phi\sin \theta=\cos (\theta-\phi)$ for the cosine function.
If we expand both $\hat F(k_{\rho},\theta)$ and $\hat f(\rho,\phi)$ in complex Fourier series, as 
$$\hat F(k_{\rho},\theta)=\sum_{n=-\infty}^\infty F_n(k_{\rho})e^{in\theta}$$
$$\hat f(\rho,\phi)=\frac{1}{2\pi}\sum_{n=-\infty}^\infty f_n(\rho)e^{in\phi+in\pi/2}$$
then $(3')$ and $(4')$ become
$$\begin{align}
\sum_{n=-\infty}^\infty F_n(k_{\rho})e^{in\theta}&=\sum_{n=-\infty}^\infty \int_0^\infty f_n(\rho) \frac{1}{2\pi}\int_0^{2\pi}  e^{in\phi+in\pi/2}\,e^{-ik_{\rho}\rho\cos(\theta-\phi)}\,d\phi\,\rho\,d\rho \\\\
&= \sum_{n=-\infty}^\infty e^{in\theta}\int_0^\infty f_n(\rho)\,J_n(k_{\rho}\rho)    \,\rho\,d\rho \tag {3''}
\end{align}$$
$$\begin{align}
\frac{1}{2\pi}\sum_{n=-\infty}^\infty f_n(\rho)e^{in\phi+in\pi/2}&=\frac{1}{(2\pi)^2}\sum_{n=-\infty}^\infty \int_0^\infty F_n(k_{\rho})\int_0^{2\pi} e^{in\theta}\,e^{ik_{\rho}\rho\cos(\theta-\phi)}\,d\theta\,k_{\rho}\,dk_{\rho}\\\\
&=\frac{1}{2\pi}\sum_{n=-\infty}^\infty e^{in\phi+in\pi/2}\int_0^\infty F_n(k_{\rho})\,J_n(k_{\rho}\rho) \,k_{\rho}\,dk_{\rho} \tag {4'}
\end{align}$$
where we used the integral representation for the $n$'th ordered Bessel Function of the First Kind given by
$$J_n(k_{\rho}\rho)=e^{-in\phi}\frac{1}{2\pi}\int_0^{2\pi}e^{ik_{\rho}\rho \cos(\theta-\phi)+in\theta-in\pi/2}\,d\theta $$
We therefore find that the Fourier coefficients satisfy the integral transform pair
$$F_n(k_{\rho})=\int_0^\infty f_n(\rho) J_n(k_{\rho}\rho)\,\rho\,d\rho \tag 5$$
$$f_n(\rho)=\int_0^\infty F_n(k_{\rho})J_n(k_{\rho}\rho)\,k_{\rho}\,dk_{\rho} \tag 6$$
If we let $f_n(\rho)=\frac{\delta(\rho-\rho')}{\rho}$ in $(5)$, then $F_n(k_{\rho})=J_n(k_{\rho}\rho')$ and we find from $(6)$ that 
$$\bbox[5px,border:2px solid #C0A000]{\frac{\delta(\rho-\rho')}{\rho}=\int_0^\infty J_n(k_{\rho}\rho')J_n(k_{\rho}\rho)\,k_{\rho}\,dk_{\rho}}$$
Similarly, if we let $F_n(k_{\rho})=\frac{\delta(k_{\rho}-k_{\rho}')}{k_{\rho}}$ in $(6)$, then $f_n(\rho)=J_n(k_{\rho}'\rho')$ and we find from $(5)$ that 
$$\bbox[5px,border:2px solid #C0A000]{\frac{\delta(k_{\rho}-k_{\rho}')}{k_{\rho}}=\int_0^\infty J_n(k_{\rho}'\rho)J_n(k_{\rho}\rho)\,\rho\,d\rho}$$
