Hankel transform of a Bessel function of different order

Question

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.

Answer 1

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

$$\begin{align} (s^2-k^2)\int_0^{\infty}J_{\nu}(kr)J_{\nu}(sr)r\,dr&=\int_0^{\infty}\left(J_{\nu}(sr)\frac{d}{dr}\left(r\frac{dJ_{\nu}(kr)}{dr}\right)-J_{\nu}(kr)\frac{d}{dr}\left(r\frac{dJ_{\nu}(sr)}{dr}\right)\right)\,dr\\\\ &=\int_0^{\infty}\frac{d}{dr}\left(rJ_{\nu}(sr)\frac{dJ_{\nu}(kr)}{dr}-rJ_{\nu}(kr)\frac{dJ_{\nu}(sr)}{dr}\right)\,dr\\\\ &=0 \end{align}$$

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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

$$\bbox[5px,border:2px solid #C0A000]{\frac{\delta(k_{\rho}-k_{\rho}')}{k_{\rho}}=\int_0^\infty J_n(k_{\rho}'\rho)J_n(k_{\rho}\rho)\,\rho\,d\rho}$$