\[\int_{0}^{\pi/2}\mathbf{H}_{\nu}\left(z\sin\theta\right)\frac{(\sin\theta)^{%
\nu+1}}{(\cos\theta)^{2\nu}}\,\mathrm{d}\theta=\frac{2^{-\nu}}{\sqrt{\pi}}%
\Gamma\left(\tfrac{1}{2}-\nu\right)z^{\nu-1}(1-\cos z),\]