B How to make this integral with initial conditions

[Safinaz]

TL;DR

I try to get the result of integration, equation ( 16 ) in this paper:

https://arxiv.org/abs/hep-ph/9905221

Hello!

The integral in equation (16), at the paper, is:

$I = r \int_{-\pi}^{\pi} e^{-2kr\phi} ~d\phi$

My integration is as the following :

$I = - \frac{1}{2 k} e^{-2kr\phi} ~|_{-\pi}^{\pi} + C$, so

$I = - \frac{1}{2 k} ( e^{-2kr\pi} -e^{2kr\pi})+ C$

Now how to use the initial conditions or how to get the result they have got?

which is

$\frac{1}{k} ( 1-e^{-2kr\pi} )$

In equation 16 there are some other factors, $M_{pl}$ and $M$ which are Planck's scales at different dimensions.

Any help is appreciated!

 

[fresh_42]

It looks as if the authors used $I=\int_{-\pi}^{\pi}=2\int_{0}^{\pi}.$

 

[Safinaz]

It looks as if the authors used $I=\int_{-\pi}^{\pi}=2\int_{0}^{\pi}.$

Ahh, it's simple like that! And they didn't consider an integration constant

 

[vanhees71]

they integrated $\exp(-2kr|\phi|)$.The modulus sign is important here.