Please fix this residue integration

Click For Summary
SUMMARY

The integral \(\int_{0}^{2\pi} \frac{d\theta}{5 - 4\sin\theta}\) evaluates to \(\frac{2\pi}{3}\). The residue integration method involves transforming the integral into a contour integral using the substitution \(z = e^{i\theta}\). The poles are identified at \(z_{1} = 2i\) and \(z_{2} = \frac{1}{2}\), leading to the calculation of the residue at \(z = 2i\), which is \(\frac{1}{6i}\). The final result is confirmed as \(-\frac{\pi}{3}\) multiplied by \(-1\), yielding the correct answer of \(\frac{2\pi}{3}\).

PREREQUISITES
  • Complex analysis fundamentals
  • Residue theorem application
  • Contour integration techniques
  • Understanding of trigonometric identities in complex form
NEXT STEPS
  • Study the residue theorem in detail
  • Practice contour integration with various functions
  • Explore the relationship between trigonometric functions and complex exponentials
  • Learn about singularities and their classification in complex analysis
USEFUL FOR

Mathematicians, physics students, and anyone interested in advanced calculus or complex analysis techniques.

geft
Messages
144
Reaction score
0
\int_{0}^{2\pi} \frac{d\theta}{5 - 4sin\theta}
= \oint_{C} \frac{dz/iz}{5 - \frac{2z}{i} - \frac{2}{iz}}
= - \oint_{C} \frac{dz}{2z^{2} - 5iz - 2}
= - \oint_{C} \frac{dz}{(4z - 8i)(49 - 2i)}
z_{1} = 2i, z_{2} = \frac{1}{2}
Res = \left|\frac{1}{4z - 2i} \right|_{z = 2i} = \frac{1}{6i}
\Rightarrow 2\pi i(\frac{1}{6i})(-1) = \frac{-\pi}{3}

The answer is 2pi/3, but I can't seem to get it.
 
Physics news on Phys.org
hi geft! :smile:

(have a pi: π :wink:)

isnt it plus 2 on the third line? :redface:
 
Oh man, I feel so stupid. This has been driving me crazy all day. Thanks a lot! :)
 

Similar threads

Complex integration is giving the wrong answer by a factor of two
  • · Replies 13 ·
Replies
13
Views
2K
Compute the residue of a function
  • · Replies 2 ·
Replies
2
Views
2K
Surface Integral of a sphere
  • · Replies 9 ·
Replies
9
Views
2K
A Definite Integral Using the Residue Theorem
  • · Replies 2 ·
Replies
2
Views
2K
Integration using residue theorem
  • · Replies 4 ·
Replies
4
Views
2K
Solve this pair of simultaneous equations involving complex numbers
  • · Replies 12 ·
Replies
12
Views
2K
Find the result of definite integration
  • · Replies 6 ·
Replies
6
Views
2K
Integration by residual theorem
  • · Replies 7 ·
Replies
7
Views
2K
Finding the nth roots of a complex number
  • · Replies 22 ·
Replies
22
Views
2K
Improper Integral Using Complex Analysis
  • · Replies 14 ·
Replies
14
Views
3K