Residue Theorem, Contour Integration, and the Cauchy Principal Value

[DivGradCurl]

Hi Folks,

I worked out a couple of problems on finding the Cauchy Principal Value, and I would like to check whether my solutions are correct and also take the opportunity to ask a couple of general questions about the residue theorem, contour integration, and the Cauchy principal value. The post looks long, but there are just a few, small questions. I did most of the work. Please help! Many thanks.

Homework Statement

Part (a) Integrate:

$${\mathcal P.V.} \int_{-\infty} ^{\infty} \frac{ \exp( -2\pi\, i\, \xi \, x ) }{4\pi ^2 \xi ^2 - k^2} \, d\xi$$

Part (b) Integrate:

$${\mathcal P.V.} \int_{-\pi} ^{\pi} \frac{e^{in\theta} }{\cos \theta} \, d\theta$$

Homework Equations

N/A

The Attempt at a Solution

Part (a) Integrate:


$${\mathcal P.V.} \int_{-\infty} ^{\infty} \frac{ \exp( -2\pi\, i\, \xi \, x ) }{4\pi ^2 \xi ^2 - k^2} \, d\xi$$

Choosing a semicircle contour of radius R and taking the limit as R goes to infinity, and also letting

$$\xi = R\, e^{i\theta}$$

gives:

$${\mathcal P.V.} \int_{-\infty} ^{\infty} \frac{ \exp( -2\pi\, i\, \xi \, x ) }{4\pi ^2 \left( \xi - \frac{k}{2\pi} \right) \left( \xi + \frac{k}{2\pi} \right) } \, d\xi = \left( \pi i \mbox{ }\sum _j \mbox{ Res}_j \right) + \lim _{R \to \infty} \int _0 ^\pi \left( d\theta i\, R \, e^{i\theta} \right) \frac{\displaystyle e^{-i\, 2\pi \, x \, R \, \cos \theta} e^{2\pi \, x \, R \, \sin \theta} }{\displaystyle 4\pi ^2 \left( R^2 e^{i\, 2\, \theta} - \frac{k^2}{4\pi^2} \right)}$$

$$= \pi i \left[ \frac{\displaystyle \left( \frac{1}{4\pi^2} \right) \, e^{-ikx} }{\displaystyle \frac{k}{\pi}} + \frac{\displaystyle \left( \frac{1}{4\pi^2} \right) \, e^{ikx} }{\displaystyle \left( -\frac{k}{\pi} \right) } \right] + 0 = \frac{1}{2k} \sin (kx)$$

As there is always a decaying exponential associated with R, regardless of whether x > 0 or x < 0; and 1/R behaviour is observed for x = 0.

3a. My Questions

- Is the above correct?
- Why do we multiply the residues by (pi * i) instead of (2*pi * i) as prescribed by the residue theorem? It seems to be the correct approach to do when the integration is along the real line, but I don't understand why.

Part (b) Integrate:

$${\mathcal P.V.} \int_{-\pi} ^{\pi} \frac{e^{in\theta} }{\cos \theta} \, d\theta$$

$$= {\mathcal P.V.} \int_{-\pi} ^{\pi} \frac{ \left( e^{i\theta} \right) ^n }{ \displaystyle \frac{1}{2} \left( e^{i\theta} + \frac{1}{e^{i\theta}} \right) } \, d\theta = {\mathcal P.V.} \mbox{ }\frac{2}{i} \, \int_{-\pi} ^{\pi} \frac{ z ^n }{ \displaystyle \frac{1}{2} \left( z + \frac{1}{z } \right)} \frac{dz}{iz} = {\mathcal P.V.} \mbox{ }\frac{2}{i} \, \int_{-\pi} ^{\pi} \frac{ z ^n }{z^2 + 1} \, dz$$

$$= \frac{2}{i} \, \left( \pi i \mbox{ }\sum _j \mbox{ Res}_j \right) + \frac{2}{i} \, \lim _{R \to \infty} \int _0 ^{\pi} \frac{ \left( R \, e^{i\theta} \right) ^n }{ \left( R \, e^{i\theta} \right) ^2 + 1 } \, \left( i\, R \, e^{i\theta} \, d\theta \right)$$

$$= \frac{2}{i} \, \left\{ \pi i \left[ \frac{i^n}{2i} + \frac{(-i)^n}{(-2i)} \right] \right\} + \frac{2}{i} \, \lim _{R \to \infty} \int _0 ^{\pi} \frac{ \left( R \, e^{i\theta} \right) ^n }{ \left( R \, e^{i\theta} \right) ^2 + 1 } \, \left( i\, R \, e^{i\theta} \, d\theta \right)$$

$$= \frac{\pi}{i} \left[ i^n - (-i)^n \right] + 0$$

3b. My Questions and comments

- Mathematica can integrate the one above; it matches my answer upon evaluation as a function of n, but it is cast in a slightly different way:

2 \[Pi] Sin[(n \[Pi])/2] + (HarmonicNumber[1/4 (-3 + n)] - HarmonicNumber[1/4 (-1 + n)]) Sin[n \[Pi]]

- How can I claim the second integral indeed goes to 0? Here is my best shot:

$$\lim _{R \to \infty} \frac{ \left( R \, e^{i\theta} \right) ^n }{ \left( R \, e^{i\theta} \right) ^2 + 1 } \, \left( i\, R \, e^{i\theta} \right) = \lim _{R \to \infty} \frac{ \left( R \, e^{i\theta} \right) ^n }{ \left( R \, e^{i\theta} \right) ^2 + 1 } \, \left( i\, R \, e^{i\theta} \right) = i \, \lim _{R \to \infty} \frac{R^{n+1} \, e^{i\theta} }{R^2 \, e^{i\, 2\theta} + 1} = i \, \lim _{R \to \infty} \frac{1}{\displaystyle \frac{R^2 \, e^{i\, 2\theta}}{R^{n+1} \, e^{i\theta}} + \frac{1}{R^{n+1} \, e^{i\theta}}} = i \, \lim _{R \to \infty} \frac{1}{\displaystyle \frac{A}{B} + C }$$

I take 3 cases for the above, namely:

i) (n+1) < 2, where I get

$$C = 0$$
$$(A/B) \to \infty$$

making the fraction go to zero

ii) (n+1) > 2, where I get

$$C = 0$$
$$(A/B) \to 0$$

making the fraction go to 1/0 = undefined

iii) (n+1) = 2, where I get

$$C = 0$$
$$A/B = e^{i(2-n)\theta}$$

making the integrand go to

$$ie^{i(n-2)\theta}$$


So, it appears that only for (n+1) < 2 the integral can be done. I might have something wrong; can someone help?

THANKS!

 

[jackmell]

First one looks ok. In regards to an integral over an indentation around a simple pole and the radius goes to zero, that's equal to $\theta i r$ where $\theta$ is the angle substended and r is the residue. So if you go around it in a half-cirlcle in the positive sense, it's just $\pi i r$. Need to check out the theorem about that in the book.

I think the second one should be converted to a closed contour. Don't know where you're getting that infinite integral but don't need it. This is how I'd do it:

$$P.V\int_{-\pi}^{\pi} \frac{e^{itn}}{\cos(t)}dt=-2i\; P.V \mathop\oint\limits_{|z|=1} \frac{z^n}{z^2+1}$$

Now, to extract the principal value from that, I'll indent the contour inward, around each pole so that the contour does not include them. That means we're going around the poles in the negative sense so that:

$$P.V \oint \frac{z^n}{z^2+1}=\pi i(r_1+r_2)$$

Finding the residues, I get:

$$\begin{align*} P.V\int_{-\pi}^{\pi} \frac{e^{itn}}{\cos(t)}dt&=-2i\; P.V \mathop\oint\limits_{|z|=1} \frac{z^n}{z^2+1}\\ &=2\pi\left[\frac{i^{n-1}}{2}\left(1-(-1)^{n-1}\right)\right]\\ &=\begin{cases} 0 && n=0\\ 0 && n \quad\text{even} \\ 2\pi i^{n-1} && n\quad\text{odd} \end{cases} \end{align*}$$

 

[DivGradCurl]

Thanks for your response, jackmell. I think I understand what you said

I think the second one should be converted to a closed contour. Don't know where you're getting that infinite integral but don't need it.

Please correct me if I'm wrong:

Because the integration limits are $-\pi$ to $\pi$, we have a counter-clockwise directed, closed contour that is an indented unit circle to exclude poles at $\pm i$. That's why we only have one integral to analyze, were its value is $\pi i \sum (\mbox{ residues at } z = \pm i)$. From that point, I get the same numerical answer as you.

 

[jackmell]

Thanks for your response, jackmell. I think I understand what you said


Please correct me if I'm wrong:

Because the integration limits are $-\pi$ to $\pi$, we have a counter-clockwise directed, closed contour that is an indented unit circle to exclude poles at $\pm i$. That's why we only have one integral to analyze, were its value is $\pi i \sum (\mbox{ residues at } z = \pm i)$. From that point, I get the same numerical answer as you.

Correct. Same concept for an equilateral triangle at the origin:

$$P.V.\mathop\oint\limits_{\begin{array}{c}\text{my}\\ \text{triangle}\end{array}} \frac{dz}{(z-1)(z+1)(z-i\sqrt{3})}=\pi/3 i(r_1+r_2+r_3)$$