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How do you solve the above integral with this one from the table of integrals?

I can't see for the life of me how to do it. I've worked others just fine.

the numerator's highest multiplier is 2n-1= 2 -1 = 1
so just 1
the bottom is 4 alpha

Ok, thanks Phi. I will try to go back and start from the beginning using what you said.

So are you saying that 2n =n and the alpha is just m/2pikbt?

n=1, not sure what you mean by 2n=n

Whoops, yes, that's what I mean.

alpha is
\[ \alpha= \frac{m}{2 K_B T} \]
and
\[\frac{1}{ \alpha}= \frac{2 K_B T}{m} \]

I am getting
\[ \frac{\sqrt{2\pi}}{2} \left( \frac{K_B T}{m}\right)^\frac{3}{2}\]

Cool fun problem! Is this Thermodynamics you're studying or Physical Chemistry?

Quantum chem!!! Its a pain!

So we are saying this is normalized then.