Exponential Integral & Incomplete Gamma function

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SUMMARY

The discussion centers on comparing the exponential integral -E_{-2k}(-m) with the Gamma function \frac{1}{m^{2k+1}}\Gamma(2k+1} as k approaches infinity. The user, Pere, seeks to understand the behavior of their difference and concludes that it tends to zero, estimating the rate of convergence with the bound \frac{e^m}{2k+1}. Additionally, Pere inquires about the definition of the exponential integral for negative real numbers and resolves the question through analytic continuation.

PREREQUISITES
  • Understanding of exponential integrals, specifically -E_{-2k}(-m)
  • Familiarity with the Gamma function, particularly \Gamma(2k+1)
  • Knowledge of limits and asymptotic behavior in mathematical analysis
  • Concept of analytic continuation in complex analysis
NEXT STEPS
  • Research the properties of the exponential integral function for negative arguments
  • Study the asymptotic behavior of the difference between exponential integrals and Gamma functions
  • Explore analytic continuation techniques in complex analysis
  • Investigate bounds and estimates in mathematical analysis
USEFUL FOR

Mathematicians, researchers in mathematical analysis, and students studying special functions and their properties.

Pere Callahan
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Hello,

I need to compare an exponential integral [tex]-E_{-2k}(-m)[/tex] -where k is a positive integer and m just a real number- to a Gamma function [tex]\frac{1}{m^{2k+1}}\Gamma(2k+1)[/tex].

I am using the notation from Mathworld here

http://mathworld.wolfram.com/ExponentialIntegral.html
http://mathworld.wolfram.com/IncompleteGammaFunction.htmlI am interested in the behaviour of their difference as [tex]k\to\infty[/tex]. It seems to tend to zero, but are there any estimates as to how fast the difference goes to zero?

Thansk for any comments. -Pere
 
Last edited:
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Maybe I ask should another question first.

How is the exponential integral function defined for real z less than zero...the integral representation clearly does not converge in that case... is it just analytic continuation or is there an explicit formula...?

Thanks

-Pere
 
Ok. Solved. I bound the difference by

[tex]\frac{e^m}{2k+1}[/tex]
 

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