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:

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

 

[Dick]

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

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

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.