Using Cauchy integral formula to compute real integral?

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Homework Help Overview

The discussion revolves around the application of Cauchy's integral formula to compute a real integral. The original poster presents a problem involving complex integration and seeks to relate the results from a complex integral to a real integral involving trigonometric functions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of Cauchy's integral formula for derivatives and how to apply the result from one integral to another. There are attempts to connect the complex integral with the real integral through trigonometric identities and the properties of complex exponentials.

Discussion Status

Some participants have provided hints and suggestions for approaching the problem, including the use of trigonometric identities and the relationship between real and imaginary parts of complex integrals. There is an ongoing exploration of how to correctly compute the integral from 0 to π based on earlier results.

Contextual Notes

There is a noted typo in the original problem statement regarding the limits of integration, which has been clarified in subsequent posts. Participants are also considering the implications of integrating over different intervals and how that affects the results.

Mixer
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Homework Statement



Compute the following integral around the path S using Cauchys integral formula for derivatives:

[itex]\int[/itex]ez / z2

Integral path S is a basic circle around origin.

Then, use the result to compute the following integral

[itex]\int[/itex] ecos (x) cos(sin (x) - x) dx from 0 to ∏

Homework Equations



Cauchy integral formula: http://en.wikipedia.org/wiki/Cauchy's_integral_formula

The Attempt at a Solution



I was able to compute the first part and got 4∏i. But I'm stuck in second part. How I'm supposed to use the result I got in first part to compute the second? Any hint for me?
 
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Mixer said:

Homework Statement



Compute the following integral around the path S using Cauchys integral formula for derivatives:

[itex]\int[/itex]ez / z2

Integral path S is a basic circle around origin.

Then, use the result to compute the following integral

[itex]\int[/itex] ecos (x) cos(sin (x) - x) dx from 0 to ∏

Homework Equations



Cauchy integral formula: http://en.wikipedia.org/wiki/Cauchy's_integral_formula

The Attempt at a Solution



I was able to compute the first part and got 4∏i. But I'm stuck in second part. How I'm supposed to use the result I got in first part to compute the second? Any hint for me?

Well, you know [itex]\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)[/itex] and you got [itex]\cos(a-b)[/itex] in there. Ok, keep that in mind.

Now, what do you get when you let [itex]z=e^{it}[/itex] in the expression:

[tex]\int \frac{e^z}{z^2}dz[/tex]

and then split the integral into a real and imaginary part?
 
Hint 1: [itex]\int_{\gamma} f(z) dz = \int^b_a f( \gamma(t) ) \gamma'(t) dt.[/itex]

Hint 2: [itex]\cos(\sin x - x) = \operatorname{Re} e^{i\sin x - ix}.[/itex]
 
Thank you very much!

I actually had a typo in my first post. So the integral I'm supposed to compute is

[itex]\int[/itex]e2 cos (t)cos(sin(t) - t) dt from 0 to ∏

I was, however, able to compute the integral you gave me to this kind of form (this is an imaginary part of the integral:

[itex]\int[/itex]e2 cos (t) cos (sin(t) - t) dt from 0 to 2∏

is equal to 4∏ (the result I got in first part)

Now, how can I get the correct value when integrating from 0 to ∏ ?
 

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