Using Cauchy integral formula to compute real integral?

Mixer
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Homework Statement



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

\intez / z2

Integral path S is a basic circle around origin.

Then, use the result to compute the following integral

\int 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:

\intez / z2

Integral path S is a basic circle around origin.

Then, use the result to compute the following integral

\int 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 \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) and you got \cos(a-b) in there. Ok, keep that in mind.

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

\int \frac{e^z}{z^2}dz

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

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

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

\inte2 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:

\inte2 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 ∏ ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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