Modified Bessel function with imaginary index is purely real?

perishingtardi
Messages
21
Reaction score
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
 
on Phys.org
I think I know what the problem is. The first equality is wrong, i.e., [itex]\left( K_{i \beta}(x) \right)^*[/itex] is not simply [itex]K_{-i\beta}(x)[/itex].
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
1
Views
11K
  • · Replies 1 ·
Replies
1
Views
2K