Integration using residue theorem

In summary, the conversation revolved around using the residue theorem to find the integral of a complex function. The process involved finding the poles and computing the residues, and there was a slight mistake in the computation of the residue which was later corrected. The final answer for the integral was $\frac{2\pi}{7}$.
  • #1
aruwin
208
0
Hi. I have to use the residue theorem to integrate f(z).
Can someone help me out? I am stuck on the factorization part.

Find the integral
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}$$

My answer:
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}=\oint_{c}^{} \,\frac{dz/iz}{25-24(\frac{1}{2}(z+\frac{1}{z}))}$$

$$=\frac{1}{i}\oint_{c}^{} \,\frac{dz}{-12z^2+25z-12}$$
 
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  • #2
aruwin said:
Hi. I have to use the residue theorem to integrate f(z).
Can someone help me out? I am stuck on the factorization part.

Find the integral
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}$$

My answer:
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}=\oint_{c}^{} \,\frac{dz/iz}{25-24(\frac{1}{2}(z+\frac{1}{z}))}$$

$$=\frac{1}{i}\oint_{c}^{} \,\frac{dz}{-12z^2+25z-12}$$

Excellent!... now find the poles of $\displaystyle f(z) = - \frac{1}{i\ (12\ z^{2} - 25\ z + 12)}$ and then compute the residues of the poles inside the unit circle...

Kind regards

$\chi$ $\sigma$
 
  • #3
Here's my continuation:

$$\frac{1}{i}\oint_{c}^{}\frac{dz}{-(4z-3)(3z-4)}=-\frac{1}{i}\oint_{c}^{}\frac{dz}{(4z-3)(3z-4)}$$

It has a singularity at z=3/4

$$Resf(z)_{|z=\frac{3}{4}|}=\lim_{{z}\to{\frac{3}{4}}}(z-\frac{3}{4})(\frac{1}{(4z-3)(3z-4)})=\frac{-7}{5}$$

By residue theorem, the integral becomes
$$2\pi{i}\frac{-1}{i}\frac{-7}{5}=\frac{14\pi}{5} $$

Is this correct? Check please.
 
  • #4
aruwin said:
Here's my continuation:

$$\frac{1}{i}\oint_{c}^{}\frac{dz}{-(4z-3)(3z-4)}=-\frac{1}{i}\oint_{c}^{}\frac{dz}{(4z-3)(3z-4)}$$

It has a singularity at z=3/4

$$Resf(z)_{|z=\frac{3}{4}|}=\lim_{{z}\to{\frac{3}{4}}}(z-\frac{3}{4})(\frac{1}{(4z-3)(3z-4)})=\frac{-7}{5}$$

By residue theorem, the integral becomes
$$2\pi{i}\frac{-1}{i}\frac{-7}{5}=\frac{14\pi}{5} $$

Is this correct? Check please.

Excellent!...

Kind regards

$\chi$ $\sigma$
 
  • #5
chisigma said:
Excellent!...

Kind regards

$\chi$ $\sigma$

I had a miscalculation. The final answer should be $\frac{2\pi}{7}$.
 
Last edited:
  • #6
aruwin said:
I had a miscalculation. The final answer should be $\frac{2\pi}{7}$.

Effectively there was a mistake in the computation of the residue...

$\displaystyle r = - \frac{1}{i}\ \lim_{z \rightarrow \frac{3}{4}} \frac{z - \frac{3}{4}}{(3 z - 4)\ (4 z - 3)} = \frac{1}{7\ i}$

Kind regards

$\chi$ $\sigma$
 

Related to Integration using residue theorem

1. What is the residue theorem?

The residue theorem is a powerful tool in complex analysis that allows for the calculation of certain types of integrals. It states that the value of a contour integral around a closed path is equal to the sum of the residues of the function at its singularities inside the contour.

2. When is the residue theorem used?

The residue theorem is typically used when integrating functions that have singularities, such as poles or branch points, on the real line. It is also commonly used in cases where the integrand is difficult to integrate using other methods.

3. How do you calculate residues?

To calculate residues, you first need to identify the singularities of the function and determine their orders. Then, you can use the formula Res(f,z0) = limz→z0 [(z-z0)nf(z)], where n is the order of the singularity. Alternatively, you can also use the Cauchy integral formula to calculate residues.

4. Can the residue theorem be used for all types of integrals?

No, the residue theorem is only applicable to certain types of integrals, namely those with singularities inside the contour. It cannot be used for integrals with no singularities or with singularities outside the contour.

5. Are there any limitations to using the residue theorem?

One limitation of the residue theorem is that it only applies to functions that are analytic, meaning they are differentiable everywhere in their domain. It also cannot be used for integrals with infinitely many singularities or for functions with essential singularities.

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