- #1

DivGradCurl

- 372

- 0

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:

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

Part (b) Integrate:

[tex]{\mathcal P.V.} \int_{-\pi} ^{\pi} \frac{e^{in\theta} }{\cos \theta} \, d\theta [/tex]

## Homework Equations

N/A

## The Attempt at a Solution

Part (a) Integrate:

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

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

[tex]\xi = R\, e^{i\theta}[/tex]

gives:

[tex] {\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)}[/tex]

[tex] = \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) [/tex]

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:

[tex]{\mathcal P.V.} \int_{-\pi} ^{\pi} \frac{e^{in\theta} }{\cos \theta} \, d\theta [/tex]

[tex]= {\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 [/tex]

[tex]= \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) [/tex]

[tex] = \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) [/tex]

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

**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:

Code:

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

[tex] \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 } [/tex]

I take 3 cases for the above, namely:

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

[tex] C = 0[/tex]

[tex] (A/B) \to \infty [/tex]

making the fraction go to zero

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

[tex] C = 0[/tex]

[tex] (A/B) \to 0 [/tex]

making the fraction go to 1/0 = undefined

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

[tex] C = 0[/tex]

[tex] A/B = e^{i(2-n)\theta}[/tex]

making the integrand go to

[tex] ie^{i(n-2)\theta} [/tex]

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

THANKS!!