Complex Integrals

  • Thread starter moo5003
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Homework Statement


Alpha(t) = e^(2ipit) for 0 <or= t <or= 1
Prove ABS(Integral over Alpha of [sin(x)/x^2)]dx) <or= 2epi

The Attempt at a Solution


Looking at the book I've come across what I believe I need to use to derive the relation, namely: ABS[Integral over alpha( f(x)dx)) <or= Cl(alpha) where
C >or= ABS(f(x)) for all x in the Image of alpha.

What I want to show then is that ABS(sin(x)/x^2) < C for all complex numbers with magnitude one, since alpha is just the unit circle.

My basic question is then, how can I show sin(x)/x^2 is bounded above given complex arguments?

Any help would be appreciated
 

Answers and Replies

  • #2
Gib Z
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Eulers Formula- [tex]e^{ix}=\cos x + i \sin x[/tex].

[tex]\sin x = \frac{e^{ix} + e^{-ix}}{2i}[/tex]. Make that substitution.
 
  • #3
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Eulers Formula- [tex]e^{ix}=\cos x + i \sin x[/tex].

[tex]\sin x = \frac{e^{ix} + e^{-ix}}{2i}[/tex]. Make that substitution.

I thought eulers formula said that sin(x) = (e^ix - e^-ix)/(2i)?
Either way, do you think proving that sin(x)/x^2 is bounded a good path to take or are you giving this substitution to work it another way?

The best simplification I've gotten using eulers and then imputting it back in is that ABS(sin(x)/x^2) = (cos(x)-1)/x^2 Even then i'm unsure where to go. I think I need to use the fact that alpha is the unit circle to somehow conclude that ABS(sin(x)/x^2) cannot exceed e, and therefore I can just say that the integral is less then or equal to the arclength (2pi) * C (e).
 
  • #4
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Alright I think I have made some progress, though I think my results are off since I didnt get what I needed explicitly.

ABS(sin(x)/x^2) = ABS(sin(x)) * ABS(1/x^2)

Since the arguments are magnitude of 1, 1/x^2 will still have magnitude of 1 and therefore ABS(1/x^2) has magnitude 1.

= ABS(sin(x)) = ABS( (e^ix - e^-ix)/(2i) )

Using triangle inequality we can break up the summation

<or= ABS( e^ix / 2i ) + ABS( e^-ix/2i )

Taking the magnitude of each of these we get:

1/2 + 1/2 = 1

Thus the integral much be less then or equal to l(alpha)C = 2pi < 2epi.

Did I make a mistake somewhere in the proof? I have very large doubts that this is completley correct, though I think the form is correct and that I'm just making a mistake somewhere in my calculation/reasoning.
 
Last edited:
  • #5
Gib Z
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Umm well to the novice mind, it seems correct. And yes you were right with your correction, I always put a little of the cosines form into the sines >.<
 

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