Efficiently Solve Gaussian Integrals with our Homework Help Guide

In summary, your proposed method of integrating the given integral using a new variable t= (p-p_0)^2 is sound, but you might need to use a bit of algebra to get the last term correct.
  • #1
andre220
75
1

Homework Statement



I need to evaluate the following integral: [tex]\sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\int\limits_{-\infty}^{\infty}p^2 e^{-32\sigma^2(p-p_0)^2/\hbar^2}\,dp[/tex]

Homework Equations



Integrals of the form: [tex] \int\limits_{-\infty}^{\infty}x^2e^{-ax^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a^3}}[/tex], taken from a table

The Attempt at a Solution



Ok so, obviously the integral provided is not of the form of the one given in the table, however, it is close. My thought was to change the integral to be such that we let some new variable [itex] t = (p-p_0)^2[/itex] so then we would have the following new integral [tex]\sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\int\limits_{-\infty}^{\infty}(t+p_0)^2 e^{-32\sigma^2 t^2/\hbar^2}\,dt[/tex]. Then I would expand it so that there are three integral that need to be evaluated i.e. (by expanding the [itex](t + p_0)^2[/itex] term.

My question is... is this mathematically sound (or allowable), I can't see a reason why this shouldn't work but I wanted to get some feedback on my proposed method. Thank you
 
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  • #2
andre220 said:

Homework Statement



I need to evaluate the following integral: [tex]\sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\int\limits_{-\infty}^{\infty}p^2 e^{-32\sigma^2(p-p_0)^2/\hbar^2}\,dp[/tex]

Homework Equations



Integrals of the form: [tex] \int\limits_{-\infty}^{\infty}x^2e^{-ax^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a^3}}[/tex], taken from a table

The Attempt at a Solution



Ok so, obviously the integral provided is not of the form of the one given in the table, however, it is close. My thought was to change the integral to be such that we let some new variable [itex] t = (p-p_0)^2[/itex] so then we would have the following new integral [tex]\sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\int\limits_{-\infty}^{\infty}(t+p_0)^2 e^{-32\sigma^2 t^2/\hbar^2}\,dt[/tex]. Then I would expand it so that there are three integral that need to be evaluated i.e. (by expanding the [itex](t + p_0)^2[/itex] term.

My question is... is this mathematically sound (or allowable), I can't see a reason why this shouldn't work but I wanted to get some feedback on my proposed method. Thank you

What you wrote is not correct because with your change of variable, ##dt \neq dp##.
But try instead ##t=p-p_0##, that will get you closer to what you want.
 
  • #3
Ok yes, I see that mistake now, so we have that [itex] t=p-p_0[/itex], then [itex]dp=dt[/itex] and so then [itex]p = t + p_0 \implies p^2 = p_0^2 + 2p_0 t + t^2[/itex] giving way to the following:
[tex] \sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\left[\,\int\limits_{-\infty}^\infty p_0^2 e^{-32\sigma^2t^2/\hbar^2}dt + \int\limits_{-\infty}^\infty 2p_0 t e^{-32\sigma^2t^2/\hbar^2}dt + \int\limits_{-\infty}^\infty t^2 e^{-32\sigma^2t^2/\hbar^2}dt\right][/tex]. Which I believe is correct? Unless I made a algebra error, or am just missing something more subtle here.
 
  • #4
andre220 said:
Ok yes, I see that mistake now, so we have that [itex] t=p-p_0[/itex], then [itex]dp=dt[/itex] and so then [itex]p = t + p_0 \implies p^2 = p_0^2 + 2p_0 t + t^2[/itex] giving way to the following:
[tex] \sqrt{\frac{2}{\pi}}\frac{\sigma}{\hbar}\left[\,\int\limits_{-\infty}^\infty p_0^2 e^{-32\sigma^2t^2/\hbar^2}dt + \int\limits_{-\infty}^\infty 2p_0 t e^{-32\sigma^2t^2/\hbar^2}dt + \int\limits_{-\infty}^\infty t^2 e^{-32\sigma^2t^2/\hbar^2}dt\right][/tex]. Which I believe is correct? Unless I made a algebra error, or am just missing something more subtle here.

I disagree with nreqd; what you wrote was absolutely correct, at least as I see it on my screen (so assuming you did not 'edit out' an error). Expanding the square ##(t + p_0)^2## and integrating term-by-term is the standard way these things are done.
 
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  • #5
I think my last reply and is the same thing I had written at first, just expanded out. I think I made an error in what I said [itex]t[/itex] should be. I wrote [itex]t=(p-p_0)^2[/itex], but what I wrote for the integral was correct, but my original statement that [itex]t=(p-p_0)^2[/itex] was not and hence he was right that [itex]dp\neq dt[/itex]. At least, that's what I gathered from the replies.
 
  • #6
andre220 said:
I think my last reply and is the same thing I had written at first, just expanded out. I think I made an error in what I said [itex]t[/itex] should be. I wrote [itex]t=(p-p_0)^2[/itex], but what I wrote for the integral was correct, but my original statement that [itex]t=(p-p_0)^2[/itex] was not and hence he was right that [itex]dp\neq dt[/itex]. At least, that's what I gathered from the replies.

Yes, this is what happened. You said ##t=(p-p_0)^2## in your OP but wrote the integral assuming ##t=p-p_0##.

You are now ready to do each integral term by term. The first 2 terms should not be too tricky, but the third term might require a little bit of a trick.
 
  • #7
The first integral is just of the form [tex]\int\limits_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}[/tex] the second integral is zero and the third should be [tex]\int\limits_{-\infty}^\infty x^2 e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}[/tex], or at least I think so.
 
  • #8
Yes. Those formulas look alright to me.
 
  • #9
Perfect, thank you all for your help.
 
  • #10
nrqed said:
What you wrote is not correct because with your change of variable, ##dt \neq dp##.
But try instead ##t=p-p_0##, that will get you closer to what you want.

In fact, if ##t = p \pm p_0## for constant ##p_0## then, indeed, ##dp = dt## is true. This is just standard change-of-variable stuff.
 
  • #11
Ray Vickson said:
In fact, if ##t = p \pm p_0## for constant ##p_0## then, indeed, ##dp = dt## is true. This is just standard change-of-variable stuff.

Yes but if you read carefully the first post, the OP wrote that he used ##t =(p-p_0)^2 ##. Then the expression written was incorrect. It's only after the fact that I realized that the integral was correct but that he actually had used ## t = p-p_0##, which is correct. So we are both right, it is just that I focused on the change of variable he wrote and did not look in details at the integral and you did the inverse.
 
  • #12
nrqed said:
Yes but if you read carefully the first post, the OP wrote that he used ##t =(p-p_0)^2 ##. Then the expression written was incorrect. It's only after the fact that I realized that the integral was correct but that he actually had used ## t = p-p_0##, which is correct. So we are both right, it is just that I focused on the change of variable he wrote and did not look in details at the integral and you did the inverse.

OK, I stand corrected.
 

Related to Efficiently Solve Gaussian Integrals with our Homework Help Guide

1. What is a Gaussian integral?

A Gaussian integral is an integral of the form ∫e^(-x^2)dx, where the integrand is a Gaussian function. This type of integral arises in many areas of mathematics, physics, and engineering, and is commonly used in probability and statistics.

2. How do I solve a Gaussian integral?

Unfortunately, there is no general formula for solving a Gaussian integral. However, there are some techniques that can be used for specific cases, such as using the substitution method or applying the properties of the Gaussian function.

3. Why are Gaussian integrals important?

Gaussian integrals are important because they have many applications in probability and statistics, as well as in physics and engineering. They also have connections to other areas of mathematics, such as complex analysis and differential equations.

4. Are there any tricks for evaluating Gaussian integrals?

Yes, there are some tricks that can be used for evaluating Gaussian integrals. For example, the substitution x = √(2t) can often simplify the integral, and the properties of the Gaussian function, such as symmetry and differentiation, can also be useful.

5. Can I use a calculator to solve a Gaussian integral?

Yes, there are some calculators that have built-in functions for evaluating Gaussian integrals. However, it is important to note that not all Gaussian integrals can be solved using a calculator, and it is still necessary to understand the concepts and techniques behind solving them.

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