How to make this integral with initial conditions

In summary, the conversation discusses an equation (16) that involves an integral, and the integration process is shown in detail. The result obtained includes factors such as Planck's scales at different dimensions. The conversation also mentions using an integration constant and the importance of the modulus sign in the integration process.
  • #1
Safinaz
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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!
 
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  • #2
It looks as if the authors used ##I=\int_{-\pi}^{\pi}=2\int_{0}^{\pi}.##
 
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  • #3
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
 
  • #4
they integrated ##\exp(-2kr|\phi|)##.The modulus sign is important here.
 

FAQ: How to make this integral with initial conditions

How do I determine the initial conditions for an integral?

The initial conditions for an integral depend on the specific problem you are trying to solve. In general, they are values or conditions that are known at the starting point of the integral. For example, if you are solving a differential equation, the initial conditions would be the values of the dependent variable and its derivatives at a specific point in time.

What is the purpose of including initial conditions in an integral?

Initial conditions are necessary in an integral to fully define the problem and find a unique solution. They provide a starting point for the integration process and help to determine the constants of integration.

How do I incorporate initial conditions into my integral?

To incorporate initial conditions into your integral, you can use them as boundary conditions or as constraints on the solution. This can be done by substituting the initial values into the integral or by using them to solve for the constants of integration.

Can I solve an integral without initial conditions?

It is possible to solve some integrals without initial conditions, but this may result in a general solution with multiple possible solutions. Including initial conditions allows for a more specific and accurate solution to be obtained.

What happens if I use incorrect initial conditions in my integral?

Using incorrect initial conditions in an integral can lead to an incorrect solution. It is important to carefully determine and use the correct initial conditions for the specific problem being solved in order to obtain an accurate solution.

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