Complex integration: which path should i choose?

[CGH]

Hi there,

this is a problem I'm having since last year, and should be "very easy", and i guess it is.
3 long years ago i took the complex analysis course, and I've used so little of it until now, that i almost forgot everything of it.

The problem is this, I have the following integration:
$$I(x)={\int_{-\infty}^{\infty} \frac{dk}{2\pi}\frac{e^{ikx}}{k^2-w^2+i\epsilon}}=\int \frac{dk}{2\pi}\frac{e^{ikx}}{(k-w+i\epsilon)(k+w-i\epsilon)}$$

taking the limit of $\epsilon$ to 0, x can be positive or negative.

If x>0, then, i use the path from $-\infty$ to $$\infty$$, then a semicircle from 0 to $\pi$ in the upper plane, and i take the residue from that enclosed region. I know that the semicirlce path is 0, because of Jordan lemma (right?), then, the only pole in that region is $k=-w+i\epsilon$, so, the result is

$$I(x>0)=i \frac{e^{-iwx}}{2w}$$

so far so good. Now the problems arrives when i take x<0, in that case, I know, Jordan lemma is not true if I take the same semicircle as before (right?), so, i use the following property:

$$e^{ikx}=e^{-ikx}+2i\sin{kx}$$

putting that in the integration, and using that the sin is odd, and the denominator even, then,

$$I(x)=\int \frac{dk}{2\pi} \frac{e^{ik(-x)}}{(k-w+i\epsilon)(k+w-i\epsilon)}$$

so, now i can use Jordan lemma and taking the semicrcle in the upper plane, so that path is 0 (right), then, using the residue theorem,

$$I(x<0)=-i \frac{e^{iwx}}{2w}$$

So, using the heaviside function,

$$I=-\frac{i}{2w}(\theta(x)e^{-iwx}+\theta(-x)e^{iwx})$$

My problem, now, is this, what if, instead taking, in the case x<0, the upper plane, i use the lower plane, and don't mind with the sine as i did, so, back to the integral

$$I(x<0)=\int \frac{dk}{2\pi} \frac{e^{ik(x)}}{(k-w+i\epsilon)(k+w-i\epsilon)}$$

then, using the lower plane (the path is from $-\infty$ to $\infty$ then from 0 to $\pi$ but from below), the Jordan lemma is again true (right?, because $e^{R\sin(x)}$ goes to 0 if x<0). Then, the only pole there is $(w-i\epsilon)$, so, i guess, the residue theorem is the same as before, just that i changed the enclosed region, then,

$$I(x<0)=i\frac{e^{iwx}}{2w}$$

if we compare the previous result i get a minus sing extra, and we all know that i good scientist miss in and even amount of sings, but i did in an odd, amount, my question is: where is my problem?

And further, what happens if i take another integral similar to the one above:

$$I(x)={\int_{-\infty}^{\infty} \frac{dk}{2\pi}\frac{e^{ikx}}{(k-i\epsilon)^2-w^2}}$$

The solution, i know, is that for x<0, I=0, if i take the lower plane it's true, but if i use

$$e^{ikx}=e^{-ikx}+2i\sin{kx}$$

and taking the upper plane i don't get that result, why?, what did i do wrong? I prefer to ask, because, as you should know, when you have and idea that something is right/wrong, unless someone tells you where's your mistake, you'll keep doing it wrong.

If you want to know, this is the time part of the Feynman propagator of a free theory, you can check the answer in any quantum field theory book, or in wikipedia: http://en.wikipedia.org/wiki/Propagator

and a final comment: if you find any mistake in my english, point it out, i'll appreciate, none of my englishspeaker friends point those thing, they think that is not "polite".

Saludos!

(and happy new year)

 

[CGH]

Hi, i guess i found my mistake:

while i was preparing the barbecue, i though "hey, cauchy is valid always, but the sing change if you change the direction of your path, like everything in your life, so, if it is anti-clockwise then

$$I=2\pi i Res()$$

but if you take the path clockwise it is

$$I=-2\pi i Res()$$

and in the latter integral,

$$I=\int \frac{dk}{2\pi}\frac{e^{ikx}}{(k-i\epsilon)^2-w^2}$$

you cannot use

$$e^{ikx}=e^{-ikx}+2i\sin(kx)$$

because

$$\int \frac{dk}{2\pi}\frac{\sin(kx)}{(k-i\epsilon)^2-w^2}$$

is not 0! (you dumb idiot!).

and so, i get the right sing, and i don't get different result for the same integral!"

So, unless i just believe that because it gives me the answer I'm looking for, that is the right answer, is it?

Saludos!

 

[mathwonk]

i love a student who is bothered by a problem three years old. you have a bright future youngster.