Solving Gaussian Integral with Integration by Parts

In summary, the formula for a general integral is given by \int^{+\infty}_{-\infty} x^{2}e^{-Ax^{2}}dx=\frac{\sqrt{\pi}}{2A^{3/2}}. However, when using integration by parts to transform the integral for a different function, there is an error in taking the derivative rather than the anti-derivative. By taking the correct anti-derivative, the appropriate form with a factor of 1/2 is obtained.
  • #1
buffordboy23
548
2
I see that the formula for this general integral is

[tex] \int^{+\infty}_{-\infty} x^{2}e^{-Ax^{2}}dx=\frac{\sqrt{\pi}}{2A^{3/2}}[/tex]

However, I am not getting this form with my function. I transformed the integral using integration by parts so that I could use another gaussian integral that I knew at the time.

[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx [/tex]

Let [tex] u = x^{2} \rightarrow du = 2x dx [/tex]

and

[tex] dv = e^{\frac{-2amx^{2}}{\hbar}}dx \rightarrow v = -\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}} [/tex]

Therefore,[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \left x^{2}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)\right|^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)2xdx[/tex]The middle term equals zero, so letting [tex]z =\left(\sqrt{2am/\hbar}\right)x \rightarrow dx= \left(\sqrt{\hbar/2am}\right)dz[/tex] gives[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \frac{\hbar}{2am}\int^{+\infty}_{-\infty}e^{\frac{-2amx^{2}}{\hbar}}\right)dx=\left(\frac{\hbar}{2am}\right)^{3/2}\int^{+\infty}_{-\infty}e^{-z^{2}}dz =\left(\frac{\hbar}{2am}\right)^{3/2}\sqrt{\pi}[/tex]

which is not in the appropriate form--missing a factor of 1/2. I can't see where I am going wrong. Any thoughts?
 
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  • #2
Take your function v and differentiate to get dv. It's not what you say it is. Use the product rule. You usually handle this problem by differentiating the integral of exp(-Ax^2) with respect to A.
 
  • #3
SOLVED

Yes, this is exactly right. Thanks dick. I forgot to consider the 'x'.
 
  • #4
buffordboy23 said:
I see that the formula for this general integral is

[tex] \int^{+\infty}_{-\infty} x^{2}e^{-Ax^{2}}dx=\frac{\sqrt{\pi}}{2A^{3/2}}[/tex]

However, I am not getting this form with my function. I transformed the integral using integration by parts so that I could use another gaussian integral that I knew at the time.

[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx [/tex]

Let [tex] u = x^{2} \rightarrow du = 2x dx [/tex]

and

[tex] dv = e^{\frac{-2amx^{2}}{\hbar}}dx \rightarrow v = -\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}} [/tex]
No, this is incorrect. It looks like you are taking the derivative rather than the anti-derivative. [itex]e^{-x^2}[/itex] does NOT have an elementary anti-derivative.

I would recommend taking u= x, [itex]dv= x e^{\frac{-2amx^{2}}{\hbar}}dx [/quote] instead.

Therefore,


[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \left x^{2}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)\right|^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)2xdx[/tex]


The middle term equals zero, so letting [tex]z =\left(\sqrt{2am/\hbar}\right)x \rightarrow dx= \left(\sqrt{\hbar/2am}\right)dz[/tex] gives


[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \frac{\hbar}{2am}\int^{+\infty}_{-\infty}e^{\frac{-2amx^{2}}{\hbar}}\right)dx=\left(\frac{\hbar}{2am}\right)^{3/2}\int^{+\infty}_{-\infty}e^{-z^{2}}dz =\left(\frac{\hbar}{2am}\right)^{3/2}\sqrt{\pi}[/tex]

which is not in the appropriate form--missing a factor of 1/2. I can't see where I am going wrong. Any thoughts?
 

1. What is the Gaussian Integral?

The Gaussian Integral is an integral that appears frequently in mathematics and physics. It is defined as the integral of the function f(x) = e-x2 over the entire real line. It has a value of √π or approximately 1.77245385091.

2. Why is solving the Gaussian Integral important?

The Gaussian Integral is important because it appears in many areas of mathematics and physics, including probability theory, statistics, and quantum mechanics. It is also used in the calculation of various physical quantities, such as the electric potential of a charged particle and the wave function in quantum mechanics.

3. What is Integration by Parts?

Integration by Parts is a method of integration that allows us to find the integral of a product of two functions. It is based on the product rule of differentiation, and involves breaking down the integral into two parts and using a specific formula to solve it.

4. How do you use Integration by Parts to solve the Gaussian Integral?

To solve the Gaussian Integral using Integration by Parts, we first rewrite the integral as the product of two functions: e-x2 and 1. Then, we use the formula ∫udv = uv - ∫vdu, where u is the first function and dv is the second function, to solve the integral step by step. This will eventually lead to a simple integral that can be easily evaluated.

5. Are there any other methods for solving the Gaussian Integral?

Yes, there are other methods for solving the Gaussian Integral, such as using the substitution method or using complex analysis techniques. However, Integration by Parts is often the most straightforward and commonly used method for solving this integral.

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