Undergrad Modified Bessel Equation

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The discussion focuses on the asymptotic behavior of modified Bessel equations for small x, particularly with complex order parameters. The user seeks clarification on interpreting the limits provided in the Wikipedia entry for Bessel functions, specifically for complex-valued α. They mention the approximation for Iα(z) and question the implications for Kα when α is complex, suggesting that α should not equal zero. The user concludes that z=0 is a regular singular point, implying that the functions should be treated as analytic. Understanding these nuances is crucial for accurately applying the modified Bessel functions in complex scenarios.
thatboi
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Hey all,
I wanted to know if anyone knew somewhere I could find the asymptotic behavior for small x (i.e x approaching 0) limit of the modified Bessel equations with complex order. The wikipedia page for Bessel functions (https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_Iα,_Kα) provides a limit when ##|z| \ll \sqrt{\alpha+1}## but in the case of complex-valued ##\alpha## I am not sure how to interpret this bound.
Any assistance would be appreciated.
 
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They give

##I_\alpha(z) \approx \frac{1}{\Gamma(\alpha+1)}\left(\frac{z}{2}\right)^\alpha##

just treat this as an analytic function of ##\alpha## for small z.
 
Paul Colby said:
They give

##I_\alpha(z) \approx \frac{1}{\Gamma(\alpha+1)}\left(\frac{z}{2}\right)^\alpha##

just treat this as an analytic function of ##\alpha## for small z.
For the ##K_{\alpha}## limits, it is defined for ##\alpha=0## and ##\alpha>0##, so for the case of complex ##\alpha##, does this just mean ##\alpha\neq 0##?
 
That’s the way I read it. ##z=0## is a regular singular point and ##\alpha## is just a parameter. So it should be analytic.
 

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