Integration of A exp( (− 3 R^2)/(2Na^2))

In summary, the conversation is about integrating an expression in order to find the normalization constant or value of A. The original poster is seeking help with integrating the expression and mentions rearranging it as A(e^-3R^2)(e0.5Na^2). However, they have not provided all the necessary details, such as the variable of integration and limits. They are also reminded to review basic properties of exponentials.
  • #1
Mic :)
48
0
Hi!
Could someone please integrate the expression (with intention of finding the normalisation constant / value of A).

Thanks a lot!
 

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  • #2
No one here will do the problem for you. You should make an attempt to solve the problem, then ask questions from there.
 
  • #3
Mic :) said:
Hi!
Could someone please integrate the expression (with intention of finding the normalisation constant / value of A).

Thanks a lot!

Your post violates PF standards and requirements. Please read the 'pinned' post by Vela, entitled 'Guidelines for students and helpers'.

In particular: (1) do not post thumbnails---type out the problem. (2) You MUST do and show some work on the problem first, before posting; this is not a homework service
 
  • #4
Excuse me; I was being hurried (thanks for the heads up, regardless). The thumbnail was posted for context.
(should I reformat and post a new thread?)

The real cheese here is integrating
A exp( (− 3 R^2)/(2Na^2))

I know that to find A / the normalisation constant, I should make the integrated expression equal to 1 (and then cube A for to get actual A for R).

It's integrating ^ that's my primary issue, any help would be great.

Thanks you!
 
  • #5
Would the first step be to arrange it as A(e^-3R^2)(e0.5Na^2) ?
 
  • #6
You need to tell us the complete problem first. You haven't said what variable you're integrating with respect to. You haven't specified the limits of integration, or perhaps it's an indefinite integral you're looking for. You also need to review basic properties of exponentials. In particular, you should know that ##e^a e^b = e^{a+b}##. Your rearrangement isn't equal to the original.

Once you get the details down, what's stopping you from evaluating it? Have you even tried to do it on your own yet?
 
  • #7
vela said:
You need to tell us the complete problem first. You haven't said what variable you're integrating with respect to. You haven't specified the limits of integration, or perhaps it's an indefinite integral you're looking for. You also need to review basic properties of exponentials. In particular, you should know that ##e^a e^b = e^{a+b}##. Your rearrangement isn't equal to the original.

Once you get the details down, what's stopping you from evaluating it? Have you even tried to do it on your own yet?

Hello!
The expression is equal to P(N,R).
Limits are infinity to - infinity.
I need to find A in terms of N and a.
 

1. What is the formula for "Integration of A exp( (− 3 R^2)/(2Na^2))"?

The formula for integrating A exp( (− 3 R^2)/(2Na^2)) is: ∫A exp( (− 3 R^2)/(2Na^2)) dR = A (πNa^2/3)^(1/2) erf((√3R)/(Na)) + C

2. What is the meaning of the variables in the formula?

The variable A represents the amplitude or magnitude of the function. R represents the variable of integration and is typically a distance or position. N represents the number of particles or molecules in the system. And a represents the characteristic length scale of the system.

3. How is this formula used in science?

This formula is commonly used in physics and chemistry to calculate the probability of finding particles or molecules at a certain distance in a system. It is also used in statistics and probability to model the distribution of random variables.

4. What is the significance of the negative exponent in the formula?

The negative exponent in the formula indicates an exponential decay, meaning that the function decreases rapidly as the distance or position increases. This is often seen in physical systems where particles tend to move away from each other.

5. Can this formula be applied to any system?

Yes, this formula can be applied to any system where the variables A, R, N, and a are relevant and the function being integrated has an exponential decay with distance or position.

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