B How to make this integral with initial conditions

Safinaz
Messages
255
Reaction score
8
TL;DR Summary
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!
 
Physics news on Phys.org
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.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top