Complexifying the integral of the secant function

In summary, the speaker is discussing their attempts to apply complexification to integration problems, specifically trying to integrate the secant of a function. They first attempted to integrate using trigonometric identities and u-substitution, but then tried to apply complexification using the expression ∫secΘdΘ = Re{∫(1/eiΘ)dΘ}. However, they were not able to get the correct answer using this method and are unsure why. They consulted a teacher who explained that the correct approach would be to use the expression cosΘ = (eiΘ+e-iΘ)/2, but did not provide an explanation. The speaker is still seeking to understand complexification and its application to integration.
  • #1
Wisc17
3
0
Since learning about being able to complexify differential equations (I am doing the MIT OCW course by Arthur Mattuck), I have tried to apply to this to particular problems in integration as well. Specifically, I have tried to integrate the secant of some function to see if it would lead to the same accepted answer as when you used trignometric identities and u substitutions. What I have done is:


∫secΘdΘ = ∫(1/cosΘ)dΘ = Re{∫(1/e)dΘ}


∫(1/e)dΘ = ∫e-iΘdΘ = -e/i = -1/(i(cosΘ+isinΘ))

= -1/(-sin+icosΘ) = -1(-sin-icosΘ)/(sin2-(-cos2Θ)) = sinΘ+icosΘ

Re(sinΘ+icosΘ) = sinΘ

I know sinΘ is not the correct answer, but I do not understand why I cannot do the math this way. In the video I watched for complexifying integrals, I watched the professor do:

∫cosΘdΘ = Re{∫edΘ}

and the professor got the correct answer doing this, so I do not understand why I cannot do this with the secant.

I have limited access to teachers right now, but I did visit one teacher at a local college. He said I cannot complexify the equation as I have done, rather that I should have done this:

cosΘ = (e+e-iΘ)/2

However, he did not explain why, and before I could ask he went off and explained how to do the integral in the "traditional way." I am not interested in the traditional way; I am trying to find another way to do the integral, and am trying to find the hole in my knowledge regarding complexifying problems.

Any help would be appreciated. Thank you.
 
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  • #2
Wisc17 said:
∫(1/cosΘ)dΘ = Re{∫(1/e)dΘ}
Why do you expect this to be true?

I see this step:
∫(1/cosΘ)dΘ = ∫(1/Re(e))dΘ

What did you do afterwards?
 

Related to Complexifying the integral of the secant function

1. What is the integral of the secant function?

The integral of the secant function is a mathematical operation that calculates the area under the curve of the secant function. It is represented by the symbol ∫sec(x) dx and is an important concept in calculus.

2. Why is complexifying the integral of the secant function important?

Complexifying the integral of the secant function allows for the integration of complex functions that cannot be solved using real numbers. It expands the scope of the integral and allows for more complex calculations and applications.

3. How is the integral of the secant function complexified?

The integral of the secant function can be complexified by introducing the concept of complex numbers and using complex analysis techniques. This involves representing the secant function as a complex function and using complex integration methods to solve it.

4. What are the practical applications of complexifying the integral of the secant function?

Complexifying the integral of the secant function has various practical applications in fields such as physics, engineering, and economics. It can be used to solve complex differential equations, calculate areas and volumes of complex shapes, and analyze electrical circuits and signals.

5. Are there any limitations to complexifying the integral of the secant function?

While complexifying the integral of the secant function expands its applications, it also introduces some limitations. It can be difficult to visualize and interpret complex functions, and the calculations can become more complex and time-consuming. Additionally, not all functions can be complexified, so it may not always be a viable solution.

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