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Complex analysis integral

  • Thread starter Physgeek64
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  • #1
247
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Homework Statement


evaluate sinx/x^4 over the unit circle

Homework Equations



Cauchys Residue theorem
##sinz=1/(2i)(z+1/z)##

The Attempt at a Solution


So we have a branch point at z=0 but its of order 4 so I can't see any direct way of using Cauchys residue theorem. Ive tried changing the sin expression to as above but simply end up with poles of order 3 and 5, which again doesn't help me.

So I tried defining a new contour essentially around the unit circle, but also enclosing the branch point by travelling back along the real axis at some small value of y. By Cauchy's residue theorem the integral along this combined contour is zero since no poles are enclosed. Redefining ##z=e^(i*theta)## for the integral along the outer circle, ##z=x+ie_0## along the path from ##x to e_0##, ##z=e_0e^(i*theta)## around the small inner circle, and ##z=x+ie_0## for the line running just above the real axis from ##e_0 to x##. But I'm not sure how to proceed from here, nor how to cary out any of the integrals.

Many thanks :)
 

Answers and Replies

  • #2
stevendaryl
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Homework Statement


evaluate sinx/x^4 over the unit circle
I think you're making this a lot more complicated than it needs to be. You know the power series for sine:

[itex]sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ... [/itex] (the general term is [itex](-1)^n \frac{x^{2n+1}}{(2n+1)!}[/itex])

So [itex]\frac{sin(x)}{x^4} = \frac{1}{x^3} - \frac{1}{6x} + \frac{x}{120} + ... [/itex] (the general term is [itex](-1)^n \frac{x^{2n-3}}{(2n+1)!}[/itex])

You can just integrate each term separately around the unit circle.

Your expression for [itex]sin(x)[/itex] has something wrong with it. It should be:

[itex]sin(x) = \frac{e^{ix} - e^{-ix}}{2i}[/itex]
 
  • #3
247
11
I think you're making this a lot more complicated than it needs to be. You know the power series for sine:

[itex]sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ... [/itex] (the general term is [itex](-1)^n \frac{x^{2n+1}}{(2n+1)!}[/itex])

So [itex]\frac{sin(x)}{x^4} = \frac{1}{x^3} - \frac{1}{6x} + \frac{x}{120} + ... [/itex] (the general term is [itex](-1)^n \frac{x^{2n-3}}{(2n+1)!}[/itex])

You can just integrate each term separately around the unit circle.

Your expression for [itex]sin(x)[/itex] has something wrong with it. It should be:

[itex]sin(x) = \frac{e^{ix} - e^{-ix}}{2i}[/itex]
Oh okay, so is it okay to use the expansion to expose the single pole?

Thank you for the reply :)

Oops, my mistake, Thank you
 
  • #4
247
11
I think you're making this a lot more complicated than it needs to be. You know the power series for sine:

[itex]sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ... [/itex] (the general term is [itex](-1)^n \frac{x^{2n+1}}{(2n+1)!}[/itex])

So [itex]\frac{sin(x)}{x^4} = \frac{1}{x^3} - \frac{1}{6x} + \frac{x}{120} + ... [/itex] (the general term is [itex](-1)^n \frac{x^{2n-3}}{(2n+1)!}[/itex])

You can just integrate each term separately around the unit circle.

Your expression for [itex]sin(x)[/itex] has something wrong with it. It should be:

[itex]sin(x) = \frac{e^{ix} - e^{-ix}}{2i}[/itex]
Just to check I got [itex] - \frac{pi*i}{3} [/itex]
 

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