# Contour Integral

## 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
Homework Helper
Let $y= e^{2it}+ w^2$

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

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
Homework Helper
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.

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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
Homework Helper
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.

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 cancelled 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
Homework Helper
I think I=2 as that factor of $2\pi i$ gets cancelled 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?

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:

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

Dick
I don't know why I didn't think to describe $w$ like that. Does this description of $I$ look OK:
$$I = \left\{ \begin{array}{lr} 2 & : \;|w|< 1\\ 0 & : \;|w|>1\\ \text{undefined} & : \; |w|=1 \end{array} \right.$$