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 $|x| \gg 1$?

  • 2
    $\begingroup$ wolframalpha.com/input/… $\endgroup$ Dec 11, 2015 at 12:44
  • $\begingroup$ they are not well defined for this case, because their taylorexpansion goes as $\sim x^{\nu}$ as x becomes very small. At least u have to specifiy the branch u are want to work with $\endgroup$
    – tired
    Dec 11, 2015 at 12:45

2 Answers 2


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.

  • $\begingroup$ It is better to cite the updated version of A&S, the NIST Handbook of Mathematical Functions or its online companion, the NIST Digital Library of Mathematical Functions. See, in particular, dlmf.nist.gov/10.34 $\endgroup$
    – Gary
    Nov 20, 2021 at 3:51
  • $\begingroup$ Olver is how I refer to the NIST Handbook of Mathematical Functions, which is why the reference to 10.34 is correct for you :-) $\endgroup$ Nov 20, 2021 at 5:19
  • $\begingroup$ Ok, but you could just refer to it right at the beginning (not mentioning A&S at all) and would not need to talk about non-integer orders separately. $\endgroup$
    – Gary
    Nov 20, 2021 at 7:54

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 $$ \begin{align} K_{\nu}(xe^{\pm \pi i}) &= \frac{\pi}{2} \frac{I_{-\nu}(xe^{\pm \pi i})-I_{\nu}(x e^{\pm \pi i})}{\sin(\nu \pi)} \\ &= \frac{\pi}{2} \frac{e^{\mp \nu \pi i}I_{-\nu}(x)-e^{\pm \nu \pi i}I_{\nu}(x)}{\sin(\nu \pi)} \\ &= \frac{\pi}{2} \frac{e^{\mp \nu \pi i}I_{- \nu}(x) -e^{\mp \nu \pi i}I_{\nu}(x)+e^{\mp \nu \pi i}I_{\nu}(x)-e^{\pm \nu \pi i}I_{\nu}(x)}{\sin(\nu \pi)} \\ &= \frac{\pi}{2} \frac{e^{\mp \nu \pi i}\left(I_{- \nu}(x) - I_{\nu}(x) \right) + \left(e^{\mp \nu \pi i} -e^{\pm \nu \pi i}\right)I_{\nu}(x)}{\sin(\nu \pi)} \\ &= e^{\mp \nu \pi i} K_{\nu}(x) \mp \pi i I_{\nu}(x). \end{align}$$


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