# Gaussian Integral

1. Aug 7, 2008

### buffordboy23

I see that the formula for this general integral is

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

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.

$$\int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx$$

Let $$u = x^{2} \rightarrow du = 2x dx$$

and

$$dv = e^{\frac{-2amx^{2}}{\hbar}}dx \rightarrow v = -\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}$$

Therefore,

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

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

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

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

Last edited: Aug 7, 2008
2. Aug 7, 2008

### Dick

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. Aug 7, 2008

### buffordboy23

SOLVED

Yes, this is exactly right. Thanks dick. I forgot to consider the 'x'.

4. Aug 7, 2008

### HallsofIvy

No, this is incorrect. It looks like you are taking the derivative rather than the anti-derivative. $e^{-x^2}$ does NOT have an elementary anti-derivative.

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