Using Cauchy integral formula to compute real integral?

In summary: Hint 1: You can take the limit as t goes to infinity.Hint 2: You could use the Laplace transform. Thank you very much for your help!In summary, I was able to compute the first part of the integral, but I'm stuck in the second part. I'm hoping someone can help me out with solving the equation for the second part.
  • #1
Mixer
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0

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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  • #2
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?
 
  • #3
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]
 
  • #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 ∏ ?
 

What is the Cauchy integral formula?

The Cauchy integral formula is a mathematical formula used to calculate the value of a complex function at a point inside its domain, given that the function is analytic over a closed contour.

How is the Cauchy integral formula used to compute real integrals?

The Cauchy integral formula can be used to compute real integrals by extending the formula to real numbers. This is done by setting the complex variable to a real value and then taking the real part of the resulting complex number.

What are the conditions for using the Cauchy integral formula?

The function must be analytic over a closed contour, the point of interest must be inside the contour, and the function must be continuous on the contour and its interior. Additionally, the contour must be simple and positively oriented.

What are the advantages of using the Cauchy integral formula to compute real integrals?

One advantage is that it can be used to evaluate integrals that are difficult or impossible to compute using other methods. It also allows for the calculation of integrals over non-standard contours, making it a versatile tool for integration.

Are there any limitations or drawbacks to using the Cauchy integral formula?

One limitation is that the function must be analytic, which means it must have a power series representation. Additionally, the contour must be closed and simple, which can limit the types of integrals that can be evaluated using this method.

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