Ill you help me?"Unraveling the Mystery of 'e': Solving an Integral

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SUMMARY

This discussion focuses on solving integrals involving the mathematical constant 'e' (approximately 2.718) in the context of quantum physics. Participants clarify the process of evaluating integrals such as ∫-∞∞ e-λu² du and the importance of substitution and boundary conditions. The conversation highlights the necessity of understanding polar coordinates for double integrals and the specific limits of integration when converting from Cartesian to polar coordinates. Key insights include the relationship between 'e' and its behavior at infinity, as well as the use of the error function for non-elementary integrals.

PREREQUISITES
  • Understanding of integral calculus, specifically definite integrals.
  • Familiarity with the mathematical constant 'e' and its properties.
  • Knowledge of polar coordinates and their application in double integrals.
  • Experience with substitution methods in integral calculus.
NEXT STEPS
  • Learn about the error function (erf) and its applications in integrals involving 'e'.
  • Study the process of converting Cartesian coordinates to polar coordinates in double integrals.
  • Explore advanced techniques for evaluating improper integrals, particularly those involving exponential functions.
  • Review the properties of the exponential function and its derivatives in calculus.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with integrals involving exponential functions and require a deeper understanding of calculus techniques.

  • #31
ah, you've calculated 4 times ∫00 dxdy …

that's 4 times ∫00π/2 drdθ, isn't it? :wink:
 
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  • #32
I think, I will not change the function when I calculate..
 

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  • #33
sorry, you're misunderstanding …

the "4" is ok :smile:

it's the "2π" in your limit which was wrong, and needed to be changed to "π/2" (in red in my last post) :wink:
 
  • #34
so, why did the function go from 0 to pi/2 ? :(

I don't understand that, on the other hand, the function could go from 0 to 2pi...

when can we see this kind of problems because I haven't figured out :(
 
  • #35
because your ∫∫ from 0 to ∞ was only over the first quadrant (x≥0, y≥0) …

that's 0 ≤ θ < π/2 :wink:

(if you'd kept to ∫∫ from -∞ to ∞, ie over all four quadrants, that would be 0 ≤ θ < 2π)
 
  • #36
thank you for everything :) this explain was clear.. by the way, which profession do you have? just curious for me:) are u student or working university? :)
 
  • #37
i am a little goldfish,

and my job is swimming around all day, and keeping an eye on things :smile:
 
  • #38
hahahaha:D, I think you don't want to talk about this.. okey , keep swimming:)
 

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