Contour integral from "QFT for the gifted amateur"

[marcom]

Hi,

Could you please help me understand the following example from page 76 of "QFT for the gifted amatur"?

I can't see how the following integral

becomes

Thanks a lot

 

[davidge]

The extra exponential term is because

$$\int_{-\infty}^{0} d|p| \ ... + \int_{0}^{\infty} d|p| \ ... \ = \int_{0}^{-\infty} d(-|p|) \ ... + \int_{0}^{\infty} d|p| \ ... \ = \ ... \ \\ = \int_{-\infty}^{\infty} d|p| |p| \exp(i |p| |x|) \bigg(\exp(it \sqrt{|p|^2 + m^2}) - \exp(-it \sqrt{|p|^2 + m^2}) \bigg)$$
Substitute $|p| = iz$ as the text says, and you get the last integral in your print.

 

[marcom]

Thanks but I didn't understand

 

[davidge]

Thanks but I didn't understand

The first integral in your post, originally with the limits $-\infty$ to $\infty$ can be re-expressed as a sum of two integrals, one from $-\infty$ to $0$ plus other one from $0$ to $\infty$.

The first one, $-\infty$ to $0$, can be re-expressed as being from $0$ to $-\infty$ as long as you change all $|p|$ by $-|p|$ in the integral. Do this to see what you get: the last integral in your print image.

 

[marcom]

I obtain:
∫d|p||p| e^{-it (√|p| 2 + m 2)}(e^i|p||x|+e^-i|p||x|)

 

[davidge]

I obtain:
∫d|p||p| e^{-it (√|p| 2 + m 2)}(e^i|p||x|+e^-i|p||x|)

That's right (you just need a minus sign in the exponential sum) . Re-write it as $\int d|p| |p| \exp(i|p||x|) \bigg(\exp \bigg(i|p| \sqrt{|p|^2+m^2} \bigg) - \exp \bigg(-i|p| \sqrt{|p|^2+m^2} \bigg) \bigg)$
and make the substitution $|p| = iz$.

 

[marcom]

OK, thanks a lot for your help!

 

[marcom]

The first one, $-\infty$ to $0$, can be re-expressed as being from $0$ to $-\infty$ as long as you change all $|p|$ by $-|p|$ in the integral. Do this to see what you get: the last integral in your print image.

Sorry to talk again about this but normally exchanging the limits of integration only introduces a minus sign
∫ _a^b=-∫_b^a
And the result wouldn't be the same, so I don't understand why the substitution p -p works.

(And to be precise in the book they calculate the integral from infinity to I am (left side) + I am to infinity (right side) because for Jordan's lemma we only have to worry about the part going up and down the cut from I am to infinity [that is equal to the integral on the real axis for Cauchy's theorem])

 

[marcom]

Perhaps because:
$\int_{-\infty}^{0} f(x)dx$=$-\int_{0}^{-\infty} f(x)dx$=$-\int_{0}^{+\infty} f(-x)d(-x)$ ?