# Complex analysis

1. Mar 3, 2009

### asi123

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

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2. Mar 3, 2009

### lurflurf

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. Mar 4, 2009

### latentcorpse

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. Mar 4, 2009

### asi123

When R goes to infinity, the integral of the semi circle goes to zero.

5. Mar 4, 2009

### latentcorpse

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. Mar 4, 2009

### asi123

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. Mar 4, 2009

### lurflurf

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: Mar 4, 2009
8. Mar 4, 2009

### latentcorpse

ok. how can you assume $n \geq 3$

9. Mar 4, 2009

### lurflurf

It is given in the problem statement.