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Contour Integration used to solved real Integrals

  • Thread starter Jufro
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  • #1
92
8

Homework Statement



Essential Mathematical Methods for the Physics Sciences Problem 15.7

Show that if f(z) has a simple zero at z0 then 1/ f(z) has a residue of 1/f'(z0). Then use this information to evaluate:

∫ sinθ/(a- sinθ) dθ, where the integral goes from -∏ to ∏.


Homework Equations


The book has further hints:
The unit circle has only one pole at z= i*a-i(a2-1)^1/2 and therefore has a residue of -i/2 *(a2 -1)-1/2.


The Attempt at a Solution



The first part is:
1/f(z) has a simple pole at z0 leaving the residue to be limz→z0 (z-z0) * 1/f(z) = 1/f'(z0) by definition of the derivative.

But using this information to do the integral is escaping me. Can anyone just help me understand how to use the hint. I don't actually need the whole step by step guide to the integral.
 

Answers and Replies

  • #2
vela
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I haven't worked it out, but I'd try using the substitution ##z=e^{i\theta}## and expressing ##\sin\theta## as ##\frac{z-1/z}{2i}##.
 
  • #3
1,796
53

Homework Statement



Essential Mathematical Methods for the Physics Sciences Problem 15.7

Show that if f(z) has a simple zero at z0 then 1/ f(z) has a residue of 1/f'(z0). Then use this information to evaluate:

∫ sinθ/(a- sinθ) dθ, where the integral goes from -∏ to ∏.


Homework Equations


The book has further hints:
The unit circle has only one pole at z= i*a-i(a2-1)^1/2 and therefore has a residue of -i/2 *(a2 -1)-1/2.


The Attempt at a Solution



The first part is:
1/f(z) has a simple pole at z0 leaving the residue to be limz→z0 (z-z0) * 1/f(z) = 1/f'(z0) by definition of the derivative.

But using this information to do the integral is escaping me. Can anyone just help me understand how to use the hint. I don't actually need the whole step by step guide to the integral.
What is a? That matters. For example if [itex]|a|<1[/itex] and real, the integral becomes a principal-valued integral.
 
  • #4
92
8
Sorry, a is real and a > 1. Rewriting sin as vela said came to the right answer, thanks.
 

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