Solving Gaussian Integral with Integration by Parts

[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)2xdxThe 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?

 

[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.

 

[buffordboy23]

SOLVED

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

 

[HallsofIvy]

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}}

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, dv= x e^{\frac{-2amx^{2}}{\hbar}}dx [/quote] instead.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Therefore,<br /> <br /> <br /> \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<br /> <br /> <br /> The middle term equals zero, so letting z =\left(\sqrt{2am/\hbar}\right)x \rightarrow dx= \left(\sqrt{\hbar/2am}\right)dz gives<br /> <br /> <br /> \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}<br /> <br /> which is not in the appropriate form--missing a factor of 1/2. I can&#039;t see where I am going wrong. Any thoughts? </div> </div> </blockquote>