Mean squared value of the Gaussian

[A Cheeky Llama]

Homework Statement

If $P(x)\propto e^{x^2/2\sigma^2}$, show that the average $\langle x^2\rangle = \sigma^2$.

Homework Equations

$\langle x\rangle = \frac {\int xP(x) \, dx} {\int P(x) \, dx}$

The Attempt at a Solution

$I = \int x^2e^{x^2/2\sigma^2} \, dx = \int (-\sigma^2x)(\frac {-x} {\sigma^2} e^{-x^2/2\sigma^2}) \, dx$
let $u = -\sigma^2x$, let $dv = \frac {-x} {\sigma^2} e^{-x^2/2\sigma^2}$
$I = -\sigma^2xe^{-x^2/2\sigma^2} + \sigma^2\int e^{-x^2/2\sigma^2} \, dx$
Therefore $\langle x^2 \rangle = \frac {-\sigma^2xe^{-x^2/2\sigma^2} + \sigma^2\int e^{-x^2/2\sigma^2} \, dx} {\int e^{-x^2/2\sigma^2} \, dx}$
$= \sigma^2 - \frac {\sigma^2xe^{-x^2/2\sigma^2}} {\int e^{-x^2/2\sigma^2} \, dx}$
At this point I'm stuck and not sure where to go. If I am to show the above, then the right hand term must equal zero, but I cannot see how this can be.
This question is just part of my individual study (from A Cavendish Quantum Mechanics Primer), I originally moved on from the question when I couldn't get it, but have kept trying to come back to it and have gotten no where.

 

[Dick]

Homework Statement

If $P(x)\propto e^{x^2/2\sigma^2}$, show that the average $\langle x^2\rangle = \sigma^2$.

Homework Equations

$\langle x\rangle = \frac {\int xP(x) \, dx} {\int P(x) \, dx}$

The Attempt at a Solution

$I = \int x^2e^{x^2/2\sigma^2} \, dx = \int (-\sigma^2x)(\frac {-x} {\sigma^2} e^{-x^2/2\sigma^2}) \, dx$
let $u = -\sigma^2x$, let $dv = \frac {-x} {\sigma^2} e^{-x^2/2\sigma^2}$
$I = -\sigma^2xe^{-x^2/2\sigma^2} + \sigma^2\int e^{-x^2/2\sigma^2} \, dx$
Therefore $\langle x^2 \rangle = \frac {-\sigma^2xe^{-x^2/2\sigma^2} + \sigma^2\int e^{-x^2/2\sigma^2} \, dx} {\int e^{-x^2/2\sigma^2} \, dx}$
$= \sigma^2 - \frac {\sigma^2xe^{-x^2/2\sigma^2}} {\int e^{-x^2/2\sigma^2} \, dx}$
At this point I'm stuck and not sure where to go. If I am to show the above, then the right hand term must equal zero, but I cannot see how this can be.
This question is just part of my individual study (from A Cavendish Quantum Mechanics Primer), I originally moved on from the question when I couldn't get it, but have kept trying to come back to it and have gotten no where.

Your left over term should be the difference between the limit of $-\sigma^2xe^{-x^2/2\sigma^2}$ as $x \to \infty$ and the limit as $x \to -\infty$. Both those limits are zero. Review what happens to that term in integration by parts.

 

[Ray Vickson]

Homework Statement

If $P(x)\propto e^{x^2/2\sigma^2}$, show that the average $\langle x^2\rangle = \sigma^2$.

Homework Equations

$\langle x\rangle = \frac {\int xP(x) \, dx} {\int P(x) \, dx}$

The Attempt at a Solution

Therefore $\langle x^2 \rangle = \frac {-\sigma^2xe^{-x^2/2\sigma^2} + \sigma^2\int e^{-x^2/2\sigma^2} \, dx} {\int e^{-x^2/2\sigma^2} \, dx}$
$= \sigma^2 - \frac {\sigma^2xe^{-x^2/2\sigma^2}} {\int e^{-x^2/2\sigma^2} \, dx}$
At this point I'm stuck and not sure where to go. If I am to show the above, then the right hand term must equal zero, but I cannot see how this can be.
This question is just part of my individual study (from A Cavendish Quantum Mechanics Primer), I originally moved on from the question when I couldn't get it, but have kept trying to come back to it and have gotten no where.

As written, your question is completely wrong: all your integrals are very badly divergent. If you had written $P(x) = K e^{-x^2/\;2/\sigma^2}$ you would have been OK, but with $+ x^2/2 \sigma^2$ in the exponent, everything blows up (unless $\sigma^2 < 0$, making $\sigma$ an imaginary number).

Anyway, you need to put limits $-\infty$ and $+\infty$ on your integrals, and then you need to remember that $P(x)$ is an even function of $x$.

 

[A Cheeky Llama]

Your left over term should be the difference between the limit of $-\sigma^2xe^{-x^2/2\sigma^2}$ as $x \to \infty$ and the limit as $x \to -\infty$. Both those limits are zero. Review what happens to that term in integration by parts.

As written, your question is completely wrong: all your integrals are very badly divergent. If you had written $P(x) = K e^{-x^2/\;2/\sigma^2}$ you would have been OK, but with $+ x^2/2 \sigma^2$ in the exponent, everything blows up (unless $\sigma^2 < 0$, making $\sigma$ an imaginary number).Anyway, you need to put limits $-\infty$ and $+\infty$ on your integrals, and then you need to remember that $P(x)$ is an even function of $x$.

Hi, sorry yeah that's a typo the question is supposed to be if $P(x)\propto e^{-x^2/2\sigma^2}$, etc

You can see further down in my working that that is what I was using, I just typed the question incorrectly on here :P

And yeah I was wondering if I was meant to put in those limits, thanks to both of you for pointing it out. I'm a little confused as to how they apply over the equation though:

Is $\int -x^2e^{x^2/2\sigma^2} \, dx$ equivalent to this: $\left. (-\sigma^2xe^{-x^2/2\sigma^2} + \sigma^2\int e^{-x^2/2\sigma^2} \, dx) \right|_{-\infty}^{\infty}$
or this: $-\sigma^2x\left. (e^{-x^2/2\sigma^2})\right|_{-\infty}^{\infty} + \sigma^2\int_{-\infty}^{\infty} \left.(e^{-x^2/2\sigma^2})\right|_{-\infty}^{\infty}\, dx$
I expect it would be the first case as I can see how that would cancel down to the answer.
Thanks for your help!

 

[Ray Vickson]

Hi, sorry yeah that's a typo the question is supposed to be if $P(x)\propto e^{-x^2/2\sigma^2}$, etc

You can see further down in my working that that is what I was using, I just typed the question incorrectly on here :P

And yeah I was wondering if I was meant to put in those limits, thanks to both of you for pointing it out. I'm a little confused as to how they apply over the equation though:

Is $\int -x^2e^{x^2/2\sigma^2} \, dx$ equivalent to this: $\left. (-\sigma^2xe^{-x^2/2\sigma^2} + \sigma^2\int e^{-x^2/2\sigma^2} \, dx) \right|_{-\infty}^{\infty}$
or this: $-\sigma^2x\left. (e^{-x^2/2\sigma^2})\right|_{-\infty}^{\infty} + \sigma^2\int_{-\infty}^{\infty} \left.(e^{-x^2/2\sigma^2})\right|_{-\infty}^{\infty}\, dx$
I expect it would be the first case as I can see how that would cancel down to the answer.
Thanks for your help!

Since $P(x)$ is an "even" function (meaning that $P(-x) = P(x)$) and $x^2$ is an even function, so is $x^2 P(x)$. Therefore,
$$\int_{-\infty}^{\infty} x^2 P(x) \, dx = 2 \int_0^{\infty} x^2 P(x) \, dx,$$
which may be a bit easier to deal with (less chance of making a simple error, etc.)

Also: $F(x) = x P(x)$ is an ""odd" function (meaning that $F(-x) = -F(x)$)---because $P$ is even and $x$ is odd. Therefore, without doing any calculations at all we have $\int_{-\infty}^{\infty} x P(x) \, dx = 0.$

 

[A Cheeky Llama]

Since $P(x)$ is an "even" function (meaning that $P(-x) = P(x)$) and $x^2$ is an even function, so is $x^2 P(x)$. Therefore,
$$\int_{-\infty}^{\infty} x^2 P(x) \, dx = 2 \int_0^{\infty} x^2 P(x) \, dx,$$
which may be a bit easier to deal with (less chance of making a simple error, etc.)

Also: $F(x) = x P(x)$ is an ""odd" function (meaning that $F(-x) = -F(x)$)---because $P$ is even and $x$ is odd. Therefore, without doing any calculations at all we have $\int_{-\infty}^{\infty} x P(x) \, dx = 0.$

Yup I can see that, but at what point in my working do I use it? I think I'm a little confused as to what order I'm meant to apply the limits. Do I apply the limits (i.e. evaluate the expression with the limits) at the very end, after simplifying everything, or do I apply them right at the very start when I'm integrating dv/dx for the by parts integration?
Thanks

 

[Ray Vickson]

Yup I can see that, but at what point in my working do I use it? I think I'm a little confused as to what order I'm meant to apply the limits. Do I apply the limits (i.e. evaluate the expression with the limits) at the very end, after simplifying everything, or do I apply them right at the very start when I'm integrating dv/dx for the by parts integration?
Thanks

Definite integration-by-parts reads as
$$\int_a^b u\, dv = \left. u v \right|_a^b - \int_a^b v \, du$$
We need to take the limits $a \to -\infty, \: b \to + \infty$ afterwards.

 

[A Cheeky Llama]

Definite integration-by-parts reads as
$$\int_a^b u\, dv = \left. u v \right|_a^b - \int_a^b v \, du$$
We need to take the limits $a \to -\infty, \: b \to + \infty$ afterwards.

Ahhh perfect that's exactly what I was looking for, thanks again for all your help :)