Modified Bessel function with imaginary index is purely real?

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SUMMARY

The modified Bessel function K_{i \beta}(x) is not purely real when both \beta and x are real, particularly for x < 0. The discussion reveals that the initial assumption that the complex conjugate of K_{i \beta}(x) equals K_{-i \beta}(x) is incorrect. Instead, the correct relationship shows that K_{i \beta}(x) can yield imaginary values under certain conditions, as confirmed by Mathematica's output. The conclusion emphasizes the need for careful consideration of the properties of Bessel functions in complex analysis.

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perishingtardi
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I'm trying to decide if the modified Bessel function K_{i \beta}(x) is purely real when \beta and x are purely real. I think that is ought to be. My reasoning is the following:

\left (K_{i \beta}(x)\right)^* = K_{-i \beta}(x) = \frac{\pi}{2} \frac{I_{i \beta}(x) - I_{-i \beta}(x)}{\sin(-i \beta\pi)} = \frac{\pi}{2} \frac{I_{i \beta}(x) - I_{-i \beta}(x)}{-\sin(i \beta\pi)} = \frac{\pi}{2} \frac{I_{-i \beta}(x) - I_{i \beta}(x)}{\sin(i \beta\pi)} = K_{i \beta}(x).

I have used here the fact that sine is an odd function and the definition of the K function in terms of the I function. So it seems that the complex conjugate of K is K itself in this case.However, Mathematica is telling me that K is imaginary if x&lt;0. Have I made a mistake somewhere? Thanks
 
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I think I know what the problem is. The first equality is wrong, i.e., \left( K_{i \beta}(x) \right)^* is not simply K_{-i\beta}(x).
 

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