Improper integral of a normal function

[happyparticle]

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.

 

[Mark44]

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.

 

[happyparticle]

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.

 

[pasmith]

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 \int f(x)\,dx is written as \int \,dx\,f(x). It is saying that <br /> \int_{-\infty}^{\infty} e^{-ax^2}\,dx = \sqrt{\frac{\pi}{a}}.

 

[happyparticle]

Alright,
I got $ab \sqrt{\pi}$

 

[Mark44]

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'.

You have misunderstood. That page is using a (IMHO nonsensical and confusing) notation in which \int f(x)\,dx is written as \int \,dx\,f(x). It is saying that <br /> \int_{-\infty}^{\infty} e^{-ax^2}\,dx = \sqrt{\frac{\pi}{a}}.

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.

 

[happyparticle]

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.

 

[WWGD]

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.