Contour Integral Homework: Evaluate I

[Ted123]

Homework Statement

The Attempt at a Solution

We can parametrise the contour \gamma (the positively oriented unit circle) by \gamma(t) = e^{it} for t \in [0, 2\pi ]

So by the definition of a contour integral

\displaystyle I = \frac{1}{2\pi i} \int^{2\pi}_0 \frac{2e^{it}}{e^{2it} + w^2} ie^{it} \; dt

\displaystyle \;\;\;= \frac{1}{\pi} \int^{2\pi}_0 \frac{e^{2it}}{e^{2it} + w^2} \; dt

How do I evaluate this?

 

[HallsofIvy]

Let y= e^{2it}+ w^2

 

[Ted123]

Let y= e^{2it}+ w^2

In fact, noticing that the numerator of the integrand of \displaystyle I = \frac{1}{2\pi i} \int^{2\pi}_0 \frac{2ie^{2it}}{e^{2it} + w^2} \; dt is the derivative of the denominator: \displaystyle I = \frac{1}{2\pi i} \left[ \log(e^{2it} + w^2) \right]^{2\pi}_0 = \frac{1}{2\pi i} [\log(1+w^2) - \log(1+w^2)] = 0

But what is this thing about the complex parameter w that the question asks about? Will the result ever not be 0?

 

[Dick]

Use the residue theorem! Don't mess with antiderivatives if there are singularities around.

 

[Ted123]

Use the residue theorem! Don't mess with antiderivatives if there are singularities around.

OK, so turning to the residue theorem:

The singularities of \displaystyle f(z) = \frac{2z}{z^2+w^2} are \pm iw.

\text{res}(f,\pm iw) = 1.

The contour \gamma is a positively oriented unit circle.

So if w \in [-1,1] then the winding number n ( \gamma , \pm iw ) =1. If w \notin [-1,1] then the winding number n ( \gamma , \pm iw ) =0.

So 2\pi i \times I = 2\pi i \left( \pm iw \times 1 \right) for w\in [-1,1] and I=0 otherwise?

 

[Dick]

OK, so turning to the residue theorem:

The singularities of \displaystyle f(z) = \frac{2z}{z^2+w^2} are \pm iw.

\text{res}(f,\pm iw) = 1.

The contour \gamma is a positively oriented unit circle.

So if w \in [-1,1] then the winding number n ( \gamma , \pm iw ) =1. If w \notin [-1,1] then the winding number n ( \gamma , \pm iw ) =0.

So 2\pi i \times I = 2\pi i \left( \pm iw \times 1 \right) for w\in [-1,1] and I=0 otherwise?

The residues aren't \pm iw. They are 1, as you said. Fix your final answer. And they want you to consider w as a complex parameter, describing it using interval notation is for real numbers. And you might want to think about the case |w|=1 separately. But that's a good start. Just needs a little fixing.

 

[Ted123]

The residues aren't \pm iw. They are 1. Fix your final answer. And they want you to consider w as a complex parameter, describing it using interval notation is for real numbers. And you might want to think about the case |w|=1 separately. But that's a good start. Just needs a little fixing.

The singularities are \pm iw ; the residues are 1 - correct?

I = (1 \times 1 + 1 \times 1) = 2 if -1 < \text{Re}(w) < 1 and -i <\text{Im}(w) < i.

and I=0 if \text{Re}(w) < -1 or \text{Re}(w) > 1 or \text{Im}(w) < -i or \text{Im}(w) > i

If |w|=1 is the winding number still 1?

 

[Dick]

The singularities are \pm iw ; the residues are 1 - correct?

I = (1 \times 1 + 1 \times 1) = 2 if -1 < \text{Re}(w) < 1 and -i <\text{Im}(w) < i.

and I=0 if \text{Re}(w) < -1 or \text{Re}(w) > 1 or \text{Im}(w) < -i or \text{Im}(w) > i

If |w|=1 is the winding number still 1?

So the integral is 4*pi*i if w inside of the contour right? The region you are describing sounds like a square to me. And the contour is round. If |w|=1 then your contour integral is singular, isn't it? The poles will be on the contour. You can still assign the integral a value if you interpret it as a Cauchy principal value, if you covered that.

 

[Ted123]

So the integral is 4*pi*i if w inside of the contour right? The region you are describing sounds like a square to me. And the contour is round. If |w|=1 then your contour integral is singular, isn't it? The poles will be on the contour. You can still assign the integral a value if you interpret it as a Cauchy principal value, if you covered that.

I think I=2 as that factor of 2\pi i gets canceled by that of the original integral. Why is the integral 'singular' if |w|=1? And yes, I know the contour is a circle but I'm having trouble describing it - I now see I was describing a square before!

 

[Dick]

I think I=2 as that factor of 2\pi i gets canceled by that of the original integral. Why is the integral 'singular' if |w|=1? And yes, I know the contour is a circle but I'm having trouble describing it - I now see I was describing a square before!

Right. The original definition cancels the 2*pi*i. The integral is singular if |w|=1 because then iw is on the unit circle. You'll get a zero in the denominator as you integrate. What's wrong with |w|<1 as a description of the interior of the unit circle?

 

[Ted123]

Right. The original definition cancels the 2*pi*i. The integral is singular if |w|=1 because then iw is on the unit circle. You'll get a zero in the denominator as you integrate. What's wrong with |w|<1 as a description of the interior of the unit circle?

I don't know why I didn't think to describe w like that. Does this description of I look OK:

I = \left\{ \begin{array}{lr} <br /> 2 &amp; : \;|w|&lt; 1\\ <br /> 0 &amp; : \;|w|&gt;1\\<br /> \text{undefined} &amp; : \; |w|=1 <br /> \end{array} <br /> \right.

 

[Dick]

I don't know why I didn't think to describe w like that. Does this description of I look OK:

I = \left\{ \begin{array}{lr} <br /> 2 &amp; : \;|w|&lt; 1\\ <br /> 0 &amp; : \;|w|&gt;1\\<br /> \text{undefined} &amp; : \; |w|=1 <br /> \end{array} <br /> \right.

Looks fine. Like I said before, if you take the principal value sense of the integral you could show you get 1 for |w|=1, but if you haven't covered that, don't worry about it.