## AmTran_Bus one year ago Integration

1. AmTran_Bus

2. AmTran_Bus

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

3. AmTran_Bus

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

4. phi

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

5. phi

and then you have sqr(pi/alpha) I would flip alpha to get 2KT/m $\frac{2 K T}{4m} \cdot \sqrt{\frac{2 KT\pi}{m} }$ and simplify that

6. AmTran_Bus

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

7. Empty

Essentially what you need to do is pattern matching. You need to bring your integral in line with this by finding what $$\alpha$$ and $$n$$ are. Don't let the constant out front of the integral distract you. So start by looking at this integral: $\int_0^\infty x^{2n} e^{-\alpha x^2} dx$ and let's make it easier on ourselves by switching out the dummy variable $$x$$ for $$\nu$$ $\int_0^\infty \nu^{2n} e^{-\alpha \nu^2} d \nu$ Ok, so we look in the integral we're given, ignoring the constant out front, that will just multiply our final answer: $\int_0^\infty \nu^{2} e^{-m \nu^2/2 k_B T} d \nu$ So now it might be easier to see what $$n$$ and $$\alpha$$ have to be for this to be true: $\int_0^\infty \nu^{2n} e^{-\alpha \nu^2} d \nu = \int_0^\infty \nu^{2} e^{-m \nu^2/2 k_B T} d \nu$ Once you have those, you just plug the values of $$n$$ and $$\alpha$$ for the solved form of the integral they have given you. Give it a try and see what you get.

8. AmTran_Bus

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

9. phi

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

10. AmTran_Bus

Whoops, yes, that's what I mean.

11. phi

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

12. phi

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

13. Empty

No, not quite, $\int_0^\infty \nu^{2n} e^{-\alpha \nu^2} d \nu = \int_0^\infty \nu^{2} e^{-m \nu^2/2 k_B T} d \nu$ What I'm saying is the terms are equal here: $$\nu^{2n} = \nu^2$$ So we will have $$2n=2$$, so $$n=1$$. Then we equate the other two terms as well: $e^{-\alpha \nu^2} =e^{-m \nu^2/2 k_B T}$ Which will give us $\alpha = m/2 k_B T$ Then we take these values and plug them into the closed form solution: $\frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2^{n+1}\alpha^n} \left( \frac{\pi}{\alpha} \right) ^{1/2}$ So the top part is the odd factorial of all the odd numbers multiplied up to 2n-1, which since n=1, that means 2n-1=2*1-1 = 1. Simple! $\frac{1}{2^{1+1}( m/2 k_B T)^1} \left( \frac{\pi}{ m/2 k_B T} \right) ^{1/2}$ Now I just plug and chug in my alpha and n values.

14. Empty

Sorry I was responding to stuff you guys kinda already figured out I guess but w/e lol I'm a bit distracted so my replies are kinda behind a step from you haha

15. AmTran_Bus

Thanks for your help all! I feel much better about it. The factorial stuff up top was throwing me for a loop.

16. Empty

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

17. AmTran_Bus

Quantum chem!!! Its a pain!

18. AmTran_Bus

So we are saying this is normalized then.