Evaluating Real Integral: Lecture 38 Example | Residue Theorem | Homework Help

[Shackleford]

Homework Statement

Use the first example in Lecture 38 to evaluate $\int_{0}^{2\pi} cost\: sint \:dt$

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Homework Equations

Residue theorem, etc.

The Attempt at a Solution

I mostly understand his work in the example. On a side note, it seems that his notes are a bit terse, but I digress.

I understand the application of the Residue Theorem around the closed path. The upper half circle includes the removable singularity z = i. He integrated around the two half circle paths using z = Re^iθ, dz =Rie^iθdz and z = re^iθ, dz = rie^iθdz substitutions. I assume that limits of integration for f(z) = f(re^iθ) are π to 0 to complete the path circumscribing the difference between the two upper half circles. For the segment integral -R to -r, why is πi in the numerator? And he equated them to segment integral r to R because of symmetry?

 

[Svein]

You are talking about two completely different things here. The problem you specify can be solved easily without referring to "Lecture 38", just remembering the formula for sin(2t).

 

[Mark44]

You are talking about two completely different things here. The problem you specify can be solved easily without referring to "Lecture 38", just remembering the formula for sin(2t).

Or even more simply, by using a very simple substitution, u = sin(t).

 

[Shackleford]

You are talking about two completely different things here. The problem you specify can be solved easily without referring to "Lecture 38", just remembering the formula for sin(2t).

Yes, I know that, but the problem says to use the first example in Lecture 38. That's why I'm first trying to fully understand the example and then apply it to the problem.

 

[Shackleford]

Or even more simply, by using a very simple substitution, u = sin(t).

Yes, of course, I know how to integrate this with real methods. I'm simply trying to do what the problem instructs.

 

[mfb]

You can transform this integral to one in the complex plane, I guess, but the most naive transformation does not lead to an analytic function.
Strange problem.

 

[Shackleford]

The best I can figure is to substitute cos θ = (1/2)[z - (1/z)], sin θ = (1/2i)[z - (1/z)], and dz = zidθ and integrate.