I'm looking for a proof of the Bessel integral identity 6.717 from Gradshteyn and Ryzhik's Tables of Integrals, Series and Products, namely
$$
\int_{-\infty}^\infty \frac{\sin(a(x+\beta))}{x^\nu(x+\beta)} J_{\nu+2n}(x) \,dx
= \pi \frac{J_{\nu+2n}(\beta)}{\beta^\nu}
$$
with $1\le a < \infty$, $n\in\mathbb{N}_0$, $\Re\nu > -\frac{3}{2}$.
I'm also wondering if there are similar Hankel-operator type integrals with sinc kernels of Bessel functions over the interval $(0, \infty)$ instead of $(-\infty, \infty)$.
Best Answer
If $\nu$ is not an integer, then $J_{\nu + 2n}(z)$ has a branch cut along the negative real axis.
I assume we're integrating on the upper side of the branch cut where $$ J_{\nu + 2n}(-x) = e^{i \pi (\nu + 2n)}J_{\nu + 2n}(x) = e^{i \pi \nu}J_{\nu + 2n}(x).$$
Then if we use the principal branch of $z^\nu$, $ z^{-\nu}J_{\nu + 2n}(z)$ is real valued on the upper side of the branch cut.
Let's integrate the function$$f(z) = \frac{e^{ia(z+ \beta)}J_{\nu+2n}(z)}{z^{\nu}(z+ \beta)}, \quad \left( a>1, \, \nu > -\frac{3}{2} \right),$$ around a contour that consists of the real axis, indented at $z=0$ and $z= -\beta$, and the semicircle above it.
The condition $a \ge 1$ is needed to ensure that the integral along the semicircle vanishes as its radius goes to infinity.
Specifically, the exponential growth of the magnitude of the Bessel function of the first kind as $\Im(z) \to +\infty$ is neutralized by the exponential decay of the magnitude of $e^{iaz}$.
Therefore, we have
$$ \begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{ia(x+ \beta)}J_{\nu+2n}(x)}{x^\nu(x+\beta)} \, \mathrm dx &= i \pi \operatorname*{Res}_{z=-\beta}f(z) \\ &= i \pi \, \frac{J_{\nu + 2 n}(-\beta)}{(-\beta)^{\nu}} \\ & = i \pi \frac{e^{i \pi v}J_{\nu + 2 n}(\beta)}{(e^{i \pi} \beta)^{\nu}} \\ &= i \pi \, \frac{J_{\nu + 2 n}(\beta)}{ \beta^{\nu}}. \end{align} $$
Equating the imaginary parts on both sides of the equation leads to the result.