\[\dfrac{\sinh z}{\pi^{2}}\int_{0}^{2\pi}\dfrac{\cos t\operatorname{me}_{\nu}'%
\left(t,h^{2}\right)\operatorname{me}_{-\nu-2m-1}\left(t,h^{2}\right)}{{\sinh}%
^{2}z+{\sin}^{2}t}\,\mathrm{d}t=(-1)^{m}\mathrm{i}h\alpha^{(1)}_{\nu,m}%
\operatorname{D}_{1}\left(\nu,\nu+2m+1,z\right),\]