Understanding the Role of Error Functions in Integrals

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Discussion Overview

The discussion revolves around the integration of an expression that leads to an imaginary error function. Participants explore the implications of evaluating the integral from negative to positive infinity and the relevance of the resulting error function.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses uncertainty about the significance of obtaining an imaginary error function from their integral and questions whether to evaluate it from the limits of -∞ to ∞ or focus on the root provided by a computational tool.
  • Another participant suggests that completing the square in the exponent is a better approach for evaluating the infinite integral, leading to a specific result involving an exponential and a square root.
  • A comment indicates that the process described is essentially taking the Fourier transform of a normal curve.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the integral or the implications of the imaginary error function, indicating multiple competing views remain.

Contextual Notes

Limitations include the potential dependence on the definitions of the error function and the Fourier transform, as well as unresolved mathematical steps in the integration process.

maverick_76
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Okay so I was integrating an expression and ended up getting an imaginary error function in the answer. I'm not sure where to go from there, I plugged it into wolfram and the root it gave me looks nice but is that worth anything to me?

The integral was being evaluated from -∞ to ∞, would I need to evaluate the error function from these limits or is the root what I'm looking for?
 
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Here is the original integral and resulting error function
9-21-2015 9-40-19 PM.jpg
9-21-2015 9-39-38 PM.jpg
 
If you want the infinite integral a better approach is to complete the square (in k) of the exponent. net result is I = e^{\frac{-x^2}{4b}}\sqrt{\frac{\pi}{b}}.

Comment: All you are doing is taking the Fourier transform of a normal curve.
 
Okay that makes sense. Thanks!
 

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