Using the Residue Theorem for Real Integrals

In summary, the conversation discusses the integration of a function using the residue theorem, but the presence of real zeros in the denominator complicates the approach. Another method, involving completing the squares, is suggested and discussed as a possible alternative.
  • #1
liorda
28
0

Homework Statement


[tex]I=\int_{-\infty}^{\infty} { dx \over {5x^2+6x+5}}[/tex]

Homework Equations


The residue theorem.

The Attempt at a Solution


I can't use the residue theorem since the denominator has real zeros. How should I solve this?
 
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  • #2
liorda said:

Homework Statement


[tex]I=\int_{-\infty}^{\infty} { dx \over {5x^2+6x+5}}[/tex]

Homework Equations


The residue theorem.

The Attempt at a Solution


I can't use the residue theorem since the denominator has real zeros. How should I solve this?
Yes you can use the residue theorem, you just need to "go around" the poles, i.e. deform your usual semi-circular contour around the poles. Have a look http://www.nhn.ou.edu/~milton/p5013/chap7.pdf" and in particular at second 7.6.
 
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  • #3
Maybe you should double-check that business about the real zeros.
 
  • #4
awkward said:
Maybe you should double-check that business about the real zeros.
Good catch, I didn't even think to look!
 
  • #5
oops. From now on, I'll never try to guess root. Quadratic equation, we meet again.

Thanks guys.
 
  • #6
There is also the quaint old "complete the squares" technique:[tex]\int_{-\infty}^{\infty} \frac{dx}{5x^2+6x+5} = \frac{1}{5} \int_{-\infty}^{\infty} \frac{dx}{ (x^2+\frac{6}{5}x+\frac{9}{25}) + (1 - \frac{9}{25} ) }[/tex]

[tex]= \frac{1}{5} \int_{-\infty}^{\infty} \frac{dx}{ (x+\frac{3}{5})^{2} + (\frac{4}{5})^{2} } . [/tex]

I believe quadratic polynomials in the denominator never require complex-analytic techniques, though you certainly aren't forbidden to use them...
 

1. What is the Residue Theorem and how is it used for real integrals?

The Residue Theorem is a powerful tool in complex analysis that allows us to evaluate certain real integrals that would be difficult or impossible to solve using traditional methods. It states that the value of a contour integral around a closed curve is equal to the sum of the residues of the singularities inside the curve, multiplied by 2πi. This allows us to convert a complex integral into a real integral that can be solved using basic calculus techniques.

2. What are singularities and how do they affect the Residue Theorem?

Singularities are points in the complex plane where a function is not analytic, meaning it is not differentiable at that point. These can be poles, where the function approaches infinity, or essential singularities, where the function has infinitely many terms in its Laurent series. The Residue Theorem relies on the fact that the function is analytic everywhere inside the contour except at the singularities, and that the singularities do not lie on the contour itself.

3. Can the Residue Theorem be used for any type of real integral?

No, the Residue Theorem can only be used for integrals that can be expressed as a contour integral around a closed curve in the complex plane. This means that the integral must have a complex variable in its integrand, and the contour must enclose all of the singularities of the function.

4. Are there any limitations or restrictions when using the Residue Theorem?

Yes, there are a few limitations and restrictions when using the Residue Theorem. Firstly, the contour must be closed, meaning it starts and ends at the same point. Additionally, the function must be analytic on and inside the contour, except at the singularities. The contour must also not pass through any singularities, as this would affect the value of the integral. Finally, the contour must be simple, meaning it does not cross over itself.

5. What are some examples of real integrals that can be solved using the Residue Theorem?

The Residue Theorem can be used to solve a variety of real integrals, including integrals involving trigonometric, exponential, and logarithmic functions. It is commonly used in physics and engineering applications to evaluate integrals involving complex numbers and functions. Some specific examples include the Gaussian integral, the Cauchy principal value integral, and integrals involving Bessel functions.

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