Evaluate Complex Integral with Residue Theorem

In summary, the conversation discusses how to evaluate the integral using the residue theorem. The speaker suggests using the substitution u^6=(x-1)/(2-x) and transforming the integral into a rational function in u. They also mention the challenge of dealing with complex numbers and suggest using z instead of x. The conversation ends with the speaker encouraging the listener to think about how to change the integral algebraically to make it easier to solve.
  • #1
paweld
255
0
How to evaluate the following integral using residue theorem:

[tex]
\int_1^2 (x+1) \sqrt[6]{\frac{x-1}{2-x}}dx
[/tex]

(The answer is [tex]\frac{31}{36}\pi [/tex])

Thanks for any help
 
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  • #2
Welcome to PF!

Welcome to PF! :smile:

Show us how far you've got, applying the residue theorem, and where you're stuck, and then we'll know how to help. :wink:
 
  • #3
That's actually a pretty tough question, in the sense that there's a lot of work involved in the calculation, but I'll get you started. You should notice that (x-1)/(2-x) goes from 0 at x=1 to infinity at x=2. That should suggest to you that you want to try the substitution u^6=(x-1)/(2-x). With some work you can use that to change the integral into a rational (even!) function in u. Since it's even you can change it into an integral over the whole real line (provided you remember to divide by 2 later). Now you just have to locate the poles etc. It looks nasty. I stopped doing the details after that. I hope this is a 'super challenge' question.
 
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  • #4
Just for habit, when dealing with complex numbers I prefer changing x's into z's. So you know you're dealing with complex stuff, such as integrals.

The problem when using residue calculus is that your contour is from 1 to 2, and in this case you residue is on 2. Which makes it break down quickly, however, if you do some nice conversions or apply some thoughtful series and you can change this =).

So, think about how you can change this algebraically to make it work for you. As a note z can be expressed in many ways.
 
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FAQ: Evaluate Complex Integral with Residue Theorem

What is the Residue Theorem?

The Residue Theorem is a powerful tool in complex analysis that allows us to evaluate certain types of complex integrals. It is based on the idea that the value of a complex integral around a closed curve is equal to the sum of the residues of the singularities inside the curve.

How do you find the residues of a complex function?

The residues of a complex function can be found by using a formula that involves taking the limit of the function as it approaches the singularity. In some cases, the residue can also be found by using the Laurent series expansion of the function.

What types of integrals can be evaluated using the Residue Theorem?

The Residue Theorem can be used to evaluate integrals of the form ∫Cf(z)dz, where C is a closed curve and f(z) is a complex function with isolated singularities inside C. This includes integrals of rational functions, trigonometric functions, and exponential functions.

Can the Residue Theorem be used to evaluate integrals over non-closed curves?

No, the Residue Theorem can only be used for integrals over closed curves. This is because the theorem relies on the Cauchy-Goursat theorem, which states that the value of a complex integral is independent of the path taken between two points as long as the points are not enclosed by a curve. Therefore, the curve must be closed in order for the theorem to be applicable.

Are there any limitations to using the Residue Theorem?

Yes, there are some limitations to using the Residue Theorem. It can only be used for integrals with isolated singularities inside the curve, and the curve must be simple and closed. Additionally, the function being integrated must be analytic (differentiable) inside the curve. If any of these conditions are not met, the Residue Theorem cannot be used to evaluate the integral.

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