Hypergeometric Limits: Analyzing p(x)

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SUMMARY

The discussion focuses on analyzing the limit of the hypergeometric function equation p(x) = (e^{x}-1)^{2i}({}_{2}F_{1}(a,b;c;e^{x}) + {}_{2}F_{1}(a+1,b+1;c+1;e^{x})) as x approaches 0. The parameters a, b, and c are complex-valued, with specific conditions on their real parts. The user seeks clarification on the interpretation of 0^{i} and the application of identities from the Digital Library of Mathematical Functions (DLMF), specifically equations (15.4.20) and (15.4.22), to properly evaluate the limit.

PREREQUISITES
  • Understanding of Hypergeometric functions, specifically the notation and properties of {}_{2}F_{1}
  • Familiarity with complex analysis, particularly limits involving complex variables
  • Knowledge of logarithmic functions and their properties in complex analysis
  • Experience with mathematical identities from the Digital Library of Mathematical Functions (DLMF)
NEXT STEPS
  • Study the properties and applications of Hypergeometric functions, focusing on {}_{2}F_{1}
  • Learn about complex limits and the implications of complex exponentiation, particularly 0^{i}
  • Review the specific identities (15.4.20) and (15.4.22) from the DLMF for their applications in limit evaluation
  • Explore advanced topics in complex analysis, such as analytic continuation and branch cuts
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Mathematicians, physicists, and graduate students specializing in complex analysis or special functions, particularly those working with Hypergeometric functions and their limits.

thatboi
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I have been working with some Hypergeometric functions whose behavior I am not quite familiar with. Suppose the equation I wish to analyze is
##p(x) = (e^{x}-1)^{2i}\left({}_{2}F_{1}(a,b;c;e^{x}) + {}_{2}F_{1}(a+1,b+1;c+1;e^{x})\right)## where ##a,b,c## are all complex valued and we have ##\Re(c-a-b)>0## and we have ##\Re((c+1)-(a+1)-(b+1))=0## but ##(a+1)+(b+1)-(c+1) = 2i##. My question is, how do I take the proper limit of ##p(x)## as ##x\rightarrow 0##. I believe the relevant identities are (15.4.20) and (15.4.22) from https://dlmf.nist.gov/15.4 but there is still the limit of figuring out what ##0^{i}## means.
Let me know if you guys have any ideas. Thanks!
 
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(e^x-1)^{2i}=e^{2\log(e^x-1)i}
So as x##\rightarrow## 0, phase becomes undermined.
 

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