# Integration using residue theory

• jameson2
In summary, the choice of contour is arbitrary as long as it encloses the poles, but the radius 1 is the most convenient choice for the original integral.
jameson2
I'm trying to evaluate the integral:
$$\int_0^{2\pi} \frac{d\theta}{24-6sin\theta}$$
using calculus of residues.

I've tried this so far:
Let $$z=e^{i\theta}$$ so

$$d\theta=\frac{dz}{iz}$$.
Also, using the exponential definition of sine,

$$sin\theta=\frac{z^2-1}{2iz}$$

This gives messy expressions for the singularities of the function, and therefore the residue I got is messy and not useful(it gives the wrong answer). The same kind of calculation is done in my book with cosine instead of sine, and it works out nicely, so I think I must be missing something in this question.

Anything in specific I should try?

Yes there is the standard substitution for these kinds of integrals. And the substitution is tan (@/2)= t. Differentiating that u get d@=2/(1+t^2)dt. Because differential of tan(@/2) is 1/2*1/cos(@/2)^2 and this is equal to1*2( 1+Tan(@/2)^2).This is the usual identity just refresh you knowledge. Then we have to express sin@ in terms of tan(@/2). U do this by following
sin(@)=2sin(@/2)cos(@/2)/1
This 1 can be written as 1= sin^2(@/2)+ cos^2(@/2).Substitute this into your fraction and then divide top and bottom by cos^2(@/2) and you get sin in terms of tan.
sin(@/2)=2t/(t^2+1).
Now u have everything and substitute these into your main integral and u should be in business.
Hope this is what u asked for, I have no idea what residue theory is:)

jameson2 said:
I'm trying to evaluate the integral:
$$\int_0^{2\pi} \frac{d\theta}{24-6sin\theta}$$
using calculus of residues.

I've tried this so far:
Let $$z=e^{i\theta}$$ so

$$d\theta=\frac{dz}{iz}$$.
Also, using the exponential definition of sine,

$$sin\theta=\frac{z^2-1}{2iz}$$

This gives messy expressions for the singularities of the function, and therefore the residue I got is messy and not useful(it gives the wrong answer). The same kind of calculation is done in my book with cosine instead of sine, and it works out nicely, so I think I must be missing something in this question.

Anything in specific I should try?

It is a simple matter of algebra. Solve the quadratic by setting the denominator to zero (i.e. after you've multiplied the 1/(i z) factor with the denominator that originally contained the sin). Then determine which of the two zeroes is inside the unit circle. You can then write the integrand in the form:

A/[(z-z1)(z-z2)]

where z1 and z2 are the two poles. The residue at z1 is then obviously A/(z1-z2).

So, you already know that the value of the integral is proportional to the reciprocal of the difference of the two zeroes. Now, if you consider the formula for the roots of the general quadratic equation and you take the difference of the two solutions, you get a simple square root. Also, you can see that taking the wrong root for the pole inside the unit circle can only yield an erroneous minus sign. You can also see this by considering the contour integral, if you take the integral to be a circle of radius R and consider the limit of R to infinity, then the integral tends to zero. But the integral is 2 pi i times the sum of both residues ion that case, implying that the two residues differ in sign.

One question I have about this is why use a unit circle as a contour? If you use a circle of radius 10 say, both roots are inside, so obviously this will give a different answer if I don't change the way I do the question. How should using a larger radius contour change how the question is done, since there can only be one answer to the original question?

jameson2 said:
One question I have about this is why use a unit circle as a contour? If you use a circle of radius 10 say, both roots are inside, so obviously this will give a different answer if I don't change the way I do the question. How should using a larger radius contour change how the question is done, since there can only be one answer to the original question?

You choose a circular contour of radius 1 when you set $z = \exp[i\theta]$, since |z| = 1. You could chose $z = r\exp[i\theta]$, with r any nonzero positive number you wanted, which would correspond to choosing a circular contour of radius r, but this would change the form of the integral you do and hence where the poles are; however, all the r dependence will cancel out in the end.

## 1. What is residue theory?

Residue theory is a mathematical concept used in complex analysis to evaluate integrals of functions with singularities.

## 2. How does residue theory work?

Residue theory works by using the Cauchy Residue theorem, which states that if a function f(z) is analytic inside and on a simple closed contour C except for a finite number of isolated singularities, then the integral of f(z) around C is equal to 2πi times the sum of the residues of f(z) at its singularities inside C.

## 3. What are singularities in residue theory?

Singularities in residue theory are points where a function is not analytic. They can be either poles, which are isolated points where the function is undefined, or branch points, where the function is multivalued.

## 4. How is residue theory used in integration?

Residue theory is used in integration by simplifying the evaluation of complex integrals that would otherwise be difficult or impossible to solve using traditional methods. It allows for the calculation of integrals using only a few key points, the singularities of the function.

## 5. What are some applications of residue theory?

Residue theory has numerous applications in physics, engineering, and other fields that deal with complex functions. It is commonly used in the evaluation of definite integrals, the calculation of residues, and the determination of Fourier series coefficients.

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