B How to make this integral with initial conditions

Safinaz
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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!
 
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It looks as if the authors used ##I=\int_{-\pi}^{\pi}=2\int_{0}^{\pi}.##
 
fresh_42 said:
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
 
they integrated ##\exp(-2kr|\phi|)##.The modulus sign is important here.
 
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