Using Cauchy integral formula to compute real integral?

Click For Summary
SUMMARY

The discussion focuses on computing integrals using Cauchy's integral formula for derivatives. The first integral, computed around a circular path S, yields a result of 4πi. The second integral, involving the expression ∫ e^cos(t) cos(sin(t) - t) dt from 0 to π, requires leveraging the result from the first integral. Participants suggest using the relationship cos(a-b) = cos(a)cos(b) + sin(a)sin(b) and transforming the integral into its real and imaginary components for further evaluation.

PREREQUISITES
  • Cauchy's integral formula
  • Complex analysis fundamentals
  • Integration techniques in calculus
  • Understanding of Euler's formula
NEXT STEPS
  • Explore the application of Cauchy's integral formula in complex analysis
  • Learn about the properties of real and imaginary parts of complex integrals
  • Study the derivation and applications of Euler's formula
  • Investigate advanced techniques for evaluating definite integrals involving trigonometric functions
USEFUL FOR

Students of complex analysis, mathematicians working on integral calculus, and anyone interested in advanced integration techniques using Cauchy's integral formula.

Mixer
Messages
38
Reaction score
0

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?
 
Physics news on Phys.org
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 ∏ ?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K