Integrate Sin(u) over [0, ∞] with Residue Calculus

In summary, the conversation discusses integrating the function \intu^-B sin(u) du from 0 to infinity, where 0<B<2. The speaker is unsure of how to approach the problem due to the condition and wonders if it can be solved using the same method as solving sin(x)/x. They mention using a systematic formula to find the residue, but this does not eliminate the need for i.
  • #1
kuahji
394
2
[tex]\int[/tex]u^-B sin(u) du, 0<B<2 integrating from 0 to infinity. What is really throwing me off is the condition, I'm not sure why it's there or really what to do with it. Can I just solve this the same way I'd solve sin(x)/x?
 
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  • #2
Had I not something strange like u^B, I'd solve the problem as follows.
[tex]\frac{1}{u^B}[/tex](u-[tex]\frac{z^3}{6}[/tex]+...)
I'd then look for C_-1, the residue. Then it's just a simple matter of 2pi(i)*res. But again, having u^B does not allow me to do this. There is a systematic formula I can apply to find the residue, but it does not allow me to eliminate i.
 

FAQ: Integrate Sin(u) over [0, ∞] with Residue Calculus

What is residue calculus?

Residue calculus is a mathematical technique used to evaluate complex integrals by finding the residues, or the values of a function at its singularities, and using them in a contour integration formula.

What is the significance of integrating Sin(u) over [0, ∞]?

Integrating Sin(u) over [0, ∞] is a common problem in mathematics and physics, as it arises in many applications such as signal processing and quantum mechanics. It is also important in understanding the behavior of periodic functions and their Fourier series.

How do you find the residues of a function?

The residues of a function can be found by finding the poles, or the points where the function becomes infinite, and then using the Laurent series expansion to find the coefficient of the term with a negative power. This coefficient is the residue at that pole.

Can residue calculus be used for any integral?

No, residue calculus can only be used for integrals that can be expressed as a complex contour integral with a finite number of singularities within the contour. It is also most effective for integrals with trigonometric or exponential functions.

What are the advantages of using residue calculus?

Residue calculus can often simplify complex integrals and make them easier to solve. It also allows for the evaluation of integrals that may be difficult or impossible to solve with other methods. Additionally, it has many applications in various fields of mathematics and physics.

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