Solving Complex Integrals: Can I Treat i as Any Other Constant?

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SUMMARY

This discussion focuses on solving integrals of the form \(\int f(v) e^{iavx} dv\) and the treatment of the imaginary unit \(i\) as a constant. Participants confirm that \(i\) can be treated as a constant during integration, allowing the integral to be expressed as a combination of real and imaginary parts using Euler's formula: \(e^{i \phi} = \cos(\phi) + i \sin(\phi)\). The final expression derived is \(w(x) = \int_{-u_0}^{u_0} i2 \pi v e^{i2 \pi vx} dv = \frac{1}{\pi x^2} \left[ 2 \pi u_0 x \cos{(2 \pi u_0 x)} - \sin{(2 \pi u_0 x)} \right]\), confirming the correctness of the approach.

PREREQUISITES
  • Understanding of complex numbers and the imaginary unit \(i\)
  • Familiarity with integral calculus and integration techniques
  • Knowledge of Euler's formula and its application in complex analysis
  • Basic understanding of trigonometric functions and their properties
NEXT STEPS
  • Study the application of Euler's formula in solving integrals involving complex exponentials
  • Learn advanced techniques in integral calculus, particularly for complex functions
  • Explore the properties of Fourier transforms and their relation to complex integrals
  • Research the implications of treating complex constants in various mathematical contexts
USEFUL FOR

Mathematicians, physics students, and engineers who are working with complex integrals and require a solid understanding of integrating functions involving complex exponentials.

Logarythmic
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How do I solve an integral of the type

\int f(v) e^{iavx} dv ?

Can I just treat i as any other constant?
 
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I'm not entirely sure I am correct about this but it seems logical to expand the complex exponent and integrate it further from there.
e^{i \phi} = \cos (\phi) + i \sin (\phi)

At a guess I would say yes, i is a constant... Just a logical guess though...
 
\int f(v) e^{iavx} dv = \int f(v) \left( \cos{avx} + i \sin{avx} \right) dv =
= \int f(v) \cos{avx} dv + i \int f(v) \sin{avx} dv

Maybe?
 
Yes, you can treat i as a constant.

Or you can use Euler's formula and write it as the sum of cos and sin, yes.
 
Using Euler's formula doesn't get rid of the i ofcourse...

Logarythmic, looks fine by me as long as you put avx in brackets ;)
 
Got it, so
w(x) = \int_{-u_0}^{u_0} i2 \pi v e^{i2 \pi vx} dv = \frac{1}{\pi x^2} \left[ 2 \pi u_0 x \cos{(2 \pi u_0 x)} - \sin{(2 \pi u_0 x)} \right]
 
I got the same, so I guess it's correct.
 

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