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Complex analysis

  • Thread starter asi123
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  • #1
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Homework Statement



Hey guys.
So, I need to calculate this integral, I uploaded what I tried to do in the pic.
But according to them, this is not the right answer, according to them, the right answer is the one I marked in red at the bottom.
Any idea where this Sin came from?

Thanks.


Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
lurflurf
Homework Helper
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One way sin may appear is from combining exponetials
sin(z)=[exp(iz)-exp(-iz)]/(2i)

Try this
integrate around a pie slice
one edge
z=t t in (0,R)
one edge
z=t*exp(pi*i/n) t in (0,R)
the rounded edge
z=R*exp(pi*i*t/n) t in (0,1)
let R become large
express the integral of the pie slice in terms of I then as 2pi*i*Res(f,exp(pi*i/(2n))
 
  • #3
1,444
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can you explain to me how you got that term to drop out to 0, when you integrated round the contour from R to -R? did you find a bound for the integrand?
 
  • #4
258
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can you explain to me how you got that term to drop out to 0, when you integrated round the contour from R to -R? did you find a bound for the integrand?
When R goes to infinity, the integral of the semi circle goes to zero.
 
  • #5
1,444
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yes. but don't you need to find a bound, say M, such that the integrand is less than or equal to M at all points on the semi circle and then say as R goes to infinity the bound goes to 0.
my question was, what did you use to bound the intergal?
 
  • #6
258
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yes. but don't you need to find a bound, say M, such that the integrand is less than or equal to M at all points on the semi circle and then say as R goes to infinity the bound goes to 0.
my question was, what did you use to bound the intergal?
Oh, I thought it's right for any semi circle integral so I didn't try to prove it.
Now I'm not sure.
 
  • #7
lurflurf
Homework Helper
2,432
132
can you explain to me how you got that term to drop out to 0, when you integrated round the contour from R to -R? did you find a bound for the integrand?
So for the curved edge
recall n>=3
|integral(curved edge)|<|R^4/(R^(2n)+1)|(2pi/n)*R<|R^5/R^(2n)|C<=C/R->0
The same bound holds for a semicircle (times n)

***It is easier to do a wedge than a semi circle as one avoids the sum of n residues only one appears.***
 
Last edited:
  • #8
1,444
0
ok. how can you assume [itex]n \geq 3[/itex]
 
  • #9
lurflurf
Homework Helper
2,432
132
ok. how can you assume [itex]n \geq 3[/itex]
It is given in the problem statement.
 

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