- #1
perishingtardi
- 21
- 1
I'm trying to decide if the modified Bessel function [tex]K_{i \beta}(x)[/tex] is purely real when [itex]\beta[/itex] and [itex]x[/itex] are purely real. I think that is ought to be. My reasoning is the following:
[tex]\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).[/tex]
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 [itex]x<0[/itex]. Have I made a mistake somewhere? Thanks
[tex]\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).[/tex]
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 [itex]x<0[/itex]. Have I made a mistake somewhere? Thanks