Evaluating integrals using the residue theorem

In summary, the conversation discusses the difficulty in evaluating simple integrals involving square roots and poles, specifically in the upper half-plane enclosing the pole at +i. The residue is found to be -i/2 and the integral should evaluate to pi/2, but the missing factor of 1/2 is not accounted for. The conversation also mentions the possibility of this being related to the Riemann sphere and its branch cuts. Additionally, the conversation suggests using the Weierstraß or tangent half-angle substitution to solve these types of integrals. Reference to a website for examples is also provided.
  • #1
sschmidt
1
0
Hi,

I have trouble evaluating simple integrals like

[tex]
\int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^2 + 2}(x^2+1)} = \frac{\pi}{2}
[/tex]
I'd like to calculate the integral closing the integration loop in the upper half-plane enclosing the pole at +i. The residue is - i / 2 and hence 2 \pi i ( - i \ 2 ) is \pi, but the integral should evaluate to \pi / 2. I just don't seem to see what is happening here. The other pole at - i shouldn't interfere and the residue at +i is not evaluated twice, so where does the missing factor 1/2 come from? Is this related to the Riemann sphere and its branch cuts? Does the contour with semicircle cross branch cuts and change the leaf?
What are the branch cuts here anyway? The same problem occurs with

[tex]
\int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^2 + 4}(x^2+1)} = \frac{2}{3} \pi i (-\frac{i}{\sqrt{3}})
[/tex]

Thank you very much and kind regards, Sven
 
Last edited:
Physics news on Phys.org
  • #2

1. What is the residue theorem and how is it used to evaluate integrals?

The residue theorem is a powerful tool in complex analysis that allows us to evaluate certain types of integrals by using the residues (singularities) of a function. It states that the integral of a function over a closed contour is equal to the sum of the residues of the function at its singularities inside the contour. This allows us to reduce complex integrals to simpler calculations involving residues.

2. What types of integrals can be evaluated using the residue theorem?

The residue theorem can be used to evaluate integrals of functions that are analytic (differentiable) inside and on a closed contour, except for a finite number of isolated singularities. These singularities can be poles, essential singularities, or branch points. The contour must also be simple, meaning it does not intersect itself.

3. How do you find the residues of a function?

To find the residues of a function, we first identify the singularities of the function inside the contour. Then, we use the Laurent series expansion of the function around each singularity to determine the coefficient of the (1/z) term. This coefficient is the residue of the function at that singularity. In some cases, we can also use the formula Res(f,z0) = lim(z→z0) [(z-z0)f(z)] to find the residue at a specific singularity z0.

4. Can the residue theorem be used to evaluate integrals along the real axis?

Yes, the residue theorem can be used to evaluate integrals along the real axis if we consider the real axis as a contour in the complex plane. However, in this case, we may need to use some additional techniques such as Jordan's lemma to handle the behavior of the integrand at infinity.

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

There are a few limitations to using the residue theorem. It can only be used for integrals of analytic functions, and the contour must be simple and closed. Additionally, the contour must contain all of the singularities of the function being integrated. If the function has infinitely many singularities, it may not be possible to evaluate the integral using the residue theorem. In some cases, the residue theorem may also give us an indeterminate result, in which case we may need to use other methods to evaluate the integral.

Similar threads

Replies
4
Views
143
  • Calculus
Replies
29
Views
444
Replies
2
Views
136
Replies
2
Views
1K
Replies
3
Views
1K
  • Calculus
Replies
8
Views
2K
  • Calculus
Replies
9
Views
766
Replies
3
Views
1K
Back
Top