Complex Analysis: Solving Integral Problem with Sin

In summary, the student attempted to solve an integral for a pie slice, but was not able to find a bound for the integral.
  • #1
asi123
258
0

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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  • #2
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
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
latentcorpse said:
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
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
latentcorpse said:
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
latentcorpse said:
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
ok. how can you assume [itex]n \geq 3[/itex]
 
  • #9
latentcorpse said:
ok. how can you assume [itex]n \geq 3[/itex]

It is given in the problem statement.
 

1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of functions of complex numbers. It involves the study of complex-valued functions, which are functions that map complex numbers to other complex numbers. Complex analysis is used to study various mathematical concepts, such as differentiation, integration, and power series, in the context of complex numbers.

2. What are the applications of complex analysis?

Complex analysis has a wide range of applications in mathematics, physics, engineering, and other fields. It is used to solve problems in fluid dynamics, electromagnetism, quantum mechanics, and signal processing. It is also used in the study of geometric and algebraic structures, such as Riemann surfaces and algebraic curves.

3. What is an integral in complex analysis?

In complex analysis, an integral is a mathematical concept that represents the area under a curve in the complex plane. It is a generalization of the concept of integration in real analysis. In complex analysis, integrals are used to calculate the values of complex-valued functions and to solve problems involving complex numbers.

4. How is the sine function used in complex analysis?

The sine function is used in complex analysis to represent the imaginary part of a complex number. It is also used to define other complex-valued functions, such as the cosine and tangent functions. In solving integral problems with sine, the sine function is used to manipulate and simplify the integrand, making it easier to solve the integral.

5. What are some techniques for solving integral problems with sine in complex analysis?

Some common techniques for solving integral problems with sine in complex analysis include using trigonometric identities, substitution, and contour integration. Trigonometric identities can be used to simplify the integrand, while substitution can be used to transform the integrand into a more manageable form. Contour integration involves using complex contour paths to evaluate the integral and is a powerful tool for solving complex integrals involving sine.

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