Improper integral of a normal function

In summary, the author is trying to solve an improper integral and is not familiar with this kind of integral. From what they found, it appears that the constants do not affect the final result.
  • #1
happyparticle
448
21
Homework Statement
Solve ##\int_{-\infty}^{\infty} (xa^3 e^{-x^2} + ab e^{-x^2}) dx##
Relevant Equations
##\int_{-\infty}^{\infty} d e^{-u^2} dx = \sqrt{\pi}##

##\int_{-\infty}^{\infty} xe^{-x^2} = 0##
I'm trying to solve an improper integral, but I'm not familiar with this kind of integral.

##\int_{-\infty}^{\infty} (xa^3 e^{-x^2} + ab e^{-x^2}) dx##
a and b are both constants.

From what I found
##\int_{-\infty}^{\infty} d e^{-u^2} dx = \sqrt{\pi}##, where d is a constant
and
##\int_{-\infty}^{\infty} xe^{-x^2} = 0##
Is it correct to say that ##\int_{-\infty}^{\infty} (xa^3 e^{-x^2} + ab e^{-x^2}) dx = \sqrt{\pi}## ?
It seems like the constants doesn't affect the final result.
 
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  • #2
happyparticle said:
Homework Statement:: Solve ##\int_{-\infty}^{\infty} (xa^3 e^{-x^2} + ab e^{-x^2}) dx##
Relevant Equations:: ##\int_{-\infty}^{\infty} d e^{-u^2} dx = \sqrt{\pi}##

##\int_{-\infty}^{\infty} xe^{-x^2} = 0##

I'm trying to solve an improper integral, but I'm not familiar with this kind of integral.

##\int_{-\infty}^{\infty} (xa^3 e^{-x^2} + ab e^{-x^2}) dx##
a and b are both constants.

From what I found
##\int_{-\infty}^{\infty} d e^{-u^2} dx = \sqrt{\pi}##, where d is a constant
and
##\int_{-\infty}^{\infty} xe^{-x^2} = 0##
Is it correct to say that ##\int_{-\infty}^{\infty} (xa^3 e^{-x^2} + ab e^{-x^2}) dx = \sqrt{\pi}## ?
It seems like the constants doesn't affect the final result.
Your first relevant equation has two errors:
  1. The exponential should be ##e^{-x^2}##
  2. From a table I found on wikipedia (https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions), ##\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac \pi a}##
Based on #2 above, the first relevant equation should read ##\int_{-\infty}^{\infty} de^{-x^2} dx = d \sqrt{ \pi }##

Regarding the integral you asked about, I would start by working with an indefinite integral, and then split them into two indefinite integrals. One of them is very simple, requiring only an ordinary substitution. The other one you can evaluate using one of your (corrected) relevant equations. After you get the antiderivatives, using limits to get the definite integrals.
 
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  • #3
From the example here : https://quantummechanics.ucsd.edu/ph130a/130_notes/node87.html
##
\int_{-\infty}^{\infty} de^{-ax^2} dx = \sqrt{\frac \pi a}
##
I assumed that ##\int_{-\infty}^{\infty} de^{-x^2} dx = \sqrt{ \pi }##
I did know how to prove it. I'll try what you told me.
 
  • #4
happyparticle said:
From the example here : https://quantummechanics.ucsd.edu/ph130a/130_notes/node87.html
##
\int_{-\infty}^{\infty} de^{-ax^2} dx = \sqrt{\frac \pi a}
##

You have misunderstood. That page is using a (IMHO nonsensical and confusing) notation in which [itex]\int f(x)\,dx[/itex] is written as [itex]\int \,dx\,f(x)[/itex]. It is saying that [tex]
\int_{-\infty}^{\infty} e^{-ax^2}\,dx = \sqrt{\frac{\pi}{a}}.[/tex]
 
  • #5
Alright,
I got ##ab \sqrt{\pi}##
 
  • #6
happyparticle said:
From the example here : https://quantummechanics.ucsd.edu/ph130a/130_notes/node87.html
##
\int_{-\infty}^{\infty} de^{-ax^2} dx = \sqrt{\frac \pi a}
##
Another error. In the web page you quoted, it says ##\int_{-\infty}^{\infty} de^{-ax^2} = \sqrt{\frac \pi a}##. Note the absence of 'dx'.
pasmith said:
You have misunderstood. That page is using a (IMHO nonsensical and confusing) notation in which [itex]\int f(x)\,dx[/itex] is written as [itex]\int \,dx\,f(x)[/itex]. It is saying that [tex]
\int_{-\infty}^{\infty} e^{-ax^2}\,dx = \sqrt{\frac{\pi}{a}}.[/tex]
I don't think that's what was meant. I initially thought that d in the expression ##de^{-ax^2}## was a constant, and my previous work was based on that assumption. However, from the post I quoted above, I now understand that 'd' is intended to mean 'differential of'. The 'dx' part included in this thread was not in the original and is incorrect.

As far as the notation ##\int dx f(x)##, I believe this is common in physics textbooks.
 
  • #7
Ah I see!
I'm so used to see dx after the function that I didn't even realize that the d was for dx.
 
  • #8
happyparticle said:
From the example here : https://quantummechanics.ucsd.edu/ph130a/130_notes/node87.html
##
\int_{-\infty}^{\infty} de^{-ax^2} dx = \sqrt{\frac \pi a}
##
I assumed that ##\int_{-\infty}^{\infty} de^{-x^2} dx = \sqrt{ \pi }##
I did know how to prove it. I'll try what you told me.
This is a slightly modified version of the integral of the Normal density function. Can be proved using Polar coordinates.
 

FAQ: Improper integral of a normal function

1. What is an improper integral of a normal function?

An improper integral of a normal function is an integral where one or both of the limits of integration are infinite or the function being integrated has an infinite discontinuity within the interval of integration.

2. How is an improper integral of a normal function evaluated?

To evaluate an improper integral of a normal function, the limits of integration are first taken to infinity or the point of discontinuity is approached. Then, the integral is evaluated using standard integration techniques.

3. What are the conditions for an improper integral of a normal function to converge?

An improper integral of a normal function will converge if the limit of the integral as the limits of integration approach infinity or the point of discontinuity is finite. Additionally, the function must be continuous or have a finite number of discontinuities within the interval of integration.

4. Can an improper integral of a normal function have a finite value if the function is unbounded?

Yes, it is possible for an improper integral of a normal function to have a finite value even if the function is unbounded. This can occur if the function has a finite number of discontinuities within the interval of integration or if the function approaches zero faster than the limits of integration approach infinity.

5. What is the significance of improper integrals of normal functions in real-world applications?

Improper integrals of normal functions are important in many areas of science, such as physics and engineering, as they allow for the evaluation of integrals that would otherwise be impossible to solve. They also have applications in statistics, where they are used to calculate probabilities and expected values of continuous random variables.

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