# §10.34 Analytic Continuation

When $m\in\Integer$,

 10.34.1 $\mathop{I_{\nu}\/}\nolimits\!\left(ze^{m\pi i}\right)=e^{m\nu\pi i}\mathop{I_{% \nu}\/}\nolimits\!\left(z\right),$
 10.34.2 $\mathop{K_{\nu}\/}\nolimits\!\left(ze^{m\pi i}\right)=e^{-m\nu\pi i}\mathop{K_% {\nu}\/}\nolimits\!\left(z\right)-\pi i\mathop{\sin\/}\nolimits\!\left(m\nu\pi% \right)\mathop{\csc\/}\nolimits\!\left(\nu\pi\right)\mathop{I_{\nu}\/}% \nolimits\!\left(z\right).$
 10.34.3 $\displaystyle\mathop{I_{\nu}\/}\nolimits\!\left(ze^{m\pi i}\right)$ $\displaystyle=(i/\pi)\left(\pm e^{m\nu\pi i}\mathop{K_{\nu}\/}\nolimits\!\left% (ze^{\pm\pi i}\right)\mp e^{(m\mp 1)\nu\pi i}\mathop{K_{\nu}\/}\nolimits\!% \left(z\right)\right),$ 10.34.4 $\displaystyle\mathop{K_{\nu}\/}\nolimits\!\left(ze^{m\pi i}\right)$ $\displaystyle=\mathop{\csc\/}\nolimits(\nu\pi)\left(\pm\mathop{\sin\/}% \nolimits(m\nu\pi)\mathop{K_{\nu}\/}\nolimits\!\left(ze^{\pm\pi i}\right)\mp% \mathop{\sin\/}\nolimits((m\mp 1)\nu\pi)\mathop{K_{\nu}\/}\nolimits\!\left(z% \right)\right).$

If $\nu=n(\in\Integer)$, then limiting values are taken in (10.34.2) and (10.34.4):

 10.34.5 $\mathop{K_{n}\/}\nolimits\!\left(ze^{m\pi i}\right)=(-1)^{mn}\mathop{K_{n}\/}% \nolimits\!\left(z\right)+(-1)^{n(m-1)-1}m\pi i\mathop{I_{n}\/}\nolimits\!% \left(z\right),$
 10.34.6 $\mathop{K_{n}\/}\nolimits\!\left(ze^{m\pi i}\right)=\pm(-1)^{n(m-1)}m\mathop{K% _{n}\/}\nolimits\!\left(ze^{\pm\pi i}\right)\mp(-1)^{nm}(m\mp 1)\mathop{K_{n}% \/}\nolimits\!\left(z\right).$

For real $\nu$,

 10.34.7 $\displaystyle\mathop{I_{\nu}\/}\nolimits\!\left(\overline{z}\right)$ $\displaystyle=\overline{\mathop{I_{\nu}\/}\nolimits\!\left(z\right)},$ $\displaystyle\mathop{K_{\nu}\/}\nolimits\!\left(\overline{z}\right)$ $\displaystyle=\overline{\mathop{K_{\nu}\/}\nolimits\!\left(z\right)}.$

For complex $\nu$ replace $\nu$ by $\overline{\nu}$ on the right-hand sides.