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Using Cauchy integral formula to compute real integral?

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data

    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 ∏

    2. Relevant equations

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

    3. 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?
     
  2. jcsd
  3. Oct 21, 2012 #2
    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?
     
  4. Oct 21, 2012 #3

    Gib Z

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    Homework Helper

    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]
     
  5. Oct 21, 2012 #4
    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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