I am working on a problem involving Bessel functions and interested in the following identity
$$ \int_0^\infty \frac{J_\nu(a x) Y_\nu(b x)-J_\nu(b x) Y_\nu(a x)} {x(J_\nu(bx)^2+Y_\nu(bx)^2)}dx = -\frac{\pi}{2}(\frac{b}{a})^\nu $$ where $J$ is the first kind Bessel function and $Y$ is the second kind Bessel function.
However, I found this identity in "Table of Integrals, Series, and Products", which has no explanation. And the reference therein provides no detail and reference neither. I hope to find more detail on how to deriving such result.
Any reference and suggestion would be greatly appreciated!
We suppose $a,b>0$. First, as the Hankel function (DLMF) \begin{equation} {H^{(1)}_{\nu}}\left(z\right)=J_{\nu}\left(z\right)+iY_{\nu}\left(z\right) \end{equation} it can be recognized that \begin{equation} \frac{J_\nu(a x) Y_\nu(b x)-J_\nu(b x) Y_\nu(a x)} {J_\nu(bx)^2+Y_\nu(bx)^2}=-\Im \left[\frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\right] \end{equation} Using the parity properties of the Hankel functions, it can be shown that \begin{align} \Re \left[\frac{H_\nu^{(1)}(axe^{i\pi})}{H_\nu^{(1)}(bxe^{i\pi})}\right]&=\Re \left[\frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\right]\\ \Im \left[\frac{H_\nu^{(1)}(axe^{i\pi})}{H_\nu^{(1)}(bxe^{i\pi})}\right]&=-\Im \left[\frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\right] \end{align} Then, \begin{align} I&=\lim_{\epsilon\to0^+}\int_\epsilon^\infty \frac{J_\nu(a x) Y_\nu(b x)-J_\nu(b x) Y_\nu(a x)} {x(J_\nu(bx)^2+Y_\nu(bx)^2)}\,dx\\ &=-\frac{1}{2i}\lim_{\epsilon\to0^+}\left[\int_{-\infty}^{-\epsilon} \frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\,\frac{dx}{x}+\int_{\epsilon}^\infty \frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\,\frac{dx}{x}\right] \end{align} We close the contour a semi-circular path, centred at the origin and avoiding the origin from above to express \begin{equation} -2iI+I_R+I_\epsilon=0 \end{equation} as no poles exist in the upper half-plane. Here, $I_R$ corresponds to the contribution of the large semi-circular path and $I_\epsilon$ to that of the small one near the origin. We observe that for $x\to \infty$ in the upper half-plane, \begin{equation} \frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\sim \sqrt{\frac{b}{a}}e^{i(a-b)x}\to 0 \end{equation} if $a>b$. Then the semi-circular path contribution to the integral vanishes. The small clock-wise half circular contribution near the origin is $$I_\epsilon=-i\pi \lim_{x\to 0} \frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}$$ It can be evaluated using the representations of $H^{(1)}_{\nu}\left(z\right)$ and of $J_\nu(z)$: \begin{align} {H^{(1)}_{\nu}}\left(z\right)&=i\csc\left(\nu\pi\right)\left(e^{-\nu\pi i}J_{\nu}\left(z\right)-J_{-\nu}\left(z\right)\right)\\ J_{\nu}\left(z\right)&=(\tfrac{1}{2}z)^{\nu}\sum_{k=0}^{\infty}(-1)^{k}\frac{( \tfrac{1}{4}z^{2})^{k}}{k!\Gamma\left(\nu+k+1\right)} \end{align} We obtain \begin{equation} I_\epsilon=\begin{cases}-i\pi\left( \frac{b}{a} \right)^\nu \text{ for } \nu>0\\ -i\pi\left( \frac{a}{b} \right)^\nu \text{ for } \nu<0 \end{cases} \end{equation} Finally \begin{equation} I=-\frac{\pi}{2}\left( \frac{b}{a} \right)^{\left|\nu\right|} \end{equation} for $0<b<a$ as tabulated in G&R (6.542), but the condition on the sign of $\nu$ is not given (at least in the 5th edition).