Modified Bessel functions with negative argument As recalled in a previous question, the modified Bessel functions of the first and second kind $I_{\nu}(x)$ and $K_{\nu}(x)$ can be obtained from $J_{\nu}(ix)$ and $N_{\nu}(ix)$: that are the Bessel functions of the first and second kind with a purely imaginary argument $z = ix$.
Note that $I_{\nu}(x)$ and $K_{\nu}(x)$ are functions of just $x$. 
The plots of $I_{\nu}(x)$ and $K_{\nu}(x)$ (like this, pp. 31-32) are always for positive values of $x$.
1) About the negative ones instead, are these functions still real-valued as with $x \gg 1$?
2) How can I plot, or where can I find a plot, of $I_{\nu}(x)$ and $K_{\nu}(x)$, with $x < 0$ and in particular $|x| \gg 1$?
 A: Very late to the party, but I found the following helpful.  Abramowitz and Stegun give the analytic continuation for $I$ and $K$ in Eq. 9.6.30 and 9.6.31 $$I_\nu(ze^{m\pi i}) = e^{m\nu\pi i}I_\nu(z), \qquad m\in\mathbb Z \\ K_\nu(ze^{m\pi i}) = e^{-m\nu\pi i}K_\nu(z) - \pi i \sin(m\nu\pi)\csc(\nu\pi)I_\nu(z), \qquad m\in\mathbb Z.$$
These results are in the Bessel Functions of Integer Order chapter, but the equations appear to hold for all orders; see, e.g., Olver 10.34.
A: The following is a way to prove the identity $$K_{\nu}(xe^{\pm \pi i}) = e^{\mp \nu \pi i} K_{\nu}(x) \mp \pi i I_{\nu}(x), \quad x >0 .$$
The above identity reflects the fact that $K_{\nu}(z)$ has a branch cut along the negative real axis.
For negative values of $x$, WolframAlpha returns the value of the the upper side of the branch cut, that is,  $$K_{\nu}(-x) = e^{- \nu \pi i}K_{\nu}(x) - \pi i I_{\nu}(x), \quad x >0. $$

For $x>0$, we have  $$ \begin{align} I_{\nu}(x  e^{\pm \pi i }) &= \sum_{m=0}^{\infty} \frac{1}{m!\Gamma(m + \nu +1)} \left(\frac{xe^{\pm \pi i}}{2} \right)^{2m + \nu} \\ &= e^{\pm \nu \pi i}\sum_{m=0}^{\infty} \frac{e^{\pm 2m \pi i}}{m!\Gamma(m + \nu +1)} \left(\frac{x}{2} \right)^{2m + \nu} \\ &=e^{\pm \nu \pi i}\sum_{m=0}^{\infty} \frac{1}{m!\Gamma(m + \nu +1)} \left(\frac{x}{2} \right)^{2m + \nu} \\ &= e^{\pm \nu \pi i} I_{\nu}(x). \end{align}$$
Then using the definition $$K_{\nu}(z) = \frac{\pi}{2}\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin(\nu \pi)} $$ we have $$ \begin{align} K_{\nu}(xe^{\pm \pi i}) &= \frac{\pi}{2} \frac{I_{-\nu}(xe^{\pm \pi i})-I_{\nu}(x e^{\pm  \pi i})}{\sin(\nu \pi)}  \\ &= \frac{\pi}{2} \frac{e^{\mp \nu \pi i}I_{-\nu}(x)-e^{\pm \nu \pi i}I_{\nu}(x)}{\sin(\nu \pi)} \\ &= \frac{\pi}{2} \frac{e^{\mp  \nu \pi i}I_{- \nu}(x) -e^{\mp  \nu \pi i}I_{\nu}(x)+e^{\mp  \nu \pi i}I_{\nu}(x)-e^{\pm \nu \pi i}I_{\nu}(x)}{\sin(\nu \pi)} \\ &= \frac{\pi}{2} \frac{e^{\mp  \nu \pi i}\left(I_{- \nu}(x) - I_{\nu}(x) \right) + \left(e^{\mp \nu \pi i} -e^{\pm  \nu \pi i}\right)I_{\nu}(x)}{\sin(\nu \pi)} \\ &= e^{\mp  \nu \pi i} K_{\nu}(x) \mp  \pi i I_{\nu}(x). \end{align}$$
