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$I_n(x)$ and $K_n(x)$ are respectively the modified Bessel function of the first and the second kind. They can be used to solve the modified Bessel differential equation.

It seems that they can be defined in multiple ways. One way is:

$$I_n(x) = -i^n J_n(ix) = i^{-n} J_n(ix) \label{a} \tag{1}$$

(where $J_n$ is the Bessel function of the first kind) and

$$K_n(x) = \frac{\pi}{2} i^{n + 1} H_n^{(1)}(ix) \label{b} \tag{2}$$

(where $H_n^{(1)}$ is the Hankel function of the first kind). $x$ is a real variable.

  1. Are $I_n(x)$ and $K_n(x)$ real functions? In other words: given the real variable $x$, are $I_n(x)$ and $K_n(x)$ always real numbers?
  2. If yes, how to prove it? The imaginary unit $i$ is in both $\ref{a}$ and $\ref{b}$ and it is seems counterintuitive that the final result is real.

Side note: any book reference, or link, with an exhaustive definition of these functions is welcome.

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1 Answer 1

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1. Yes: if $x \in \mathbb{R}$, they are always real functions. In fact, they have been obtained ad hoc to formulate a general, real solution for the modified Bessel differential equation:

$$f(x) = AI_{\nu}(x) + BK_{\nu}(x)$$

It is real, if $A, B \in \mathbb{R}$.

2. The modified Bessel differential equation is linear: if $J_{\nu}(ix)$ is a solution, $CJ_{\nu}(ix)$, $C \in \mathbb{C}$ is a solution, too. Observing the series definition of $J_{\nu}(ix)$, the arbitrary choice $C = i^{-\nu}$ will generate a real function of the variable $x$ only: $I_{\nu}(x) = i^{-\nu} J_{\nu}(ix)$ is the modified Bessel function of the first kind. The modified Bessel function of the second kind $K_{\nu}(x)$ is also defined as

$$K_{\nu}(x) = \pi \frac{\left[ I_{-\nu}(x) - I_{\nu} (x) \right]}{2 \sin (\nu \pi)}$$

for $\nu \in \mathbb{C} \setminus \mathbb{Z}$ , and as the limit of the above expression for $\nu \to n \in \mathbb{Z}$. If $\nu$ is real, being $I_{\nu}(x)$ and $I_{-\nu}(x)$ real functions, $K_{\nu}(x)$ will be real, too.

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