I've got this expression $$\int_0^{\infty} 4 \pi v^2 C e^{-\frac{mv^{2}}{2kT}} \, dv$$ from 0 to ∞.
I've tried everything from by parts to tabular but couldn't get anywhere. Is this even integrable?
Asked
Active
Viewed 100 times
0
-
Could you edit the formula ? Thanks – Claude Leibovici Jun 11 '15 at 10:13
-
Trying to. Using a cell phone -_- – Foon Jun 11 '15 at 10:15
-
You should also say what is the variable over which we are integrating. – Crostul Jun 11 '15 at 10:19
-
Why wouldn't parts work? $$v e^{-a v^2}dv = -\frac1{2 a} d\left (e^{-a v^2}\right )$$ – Ron Gordon Jun 11 '15 at 10:24
-
Am I supposed to do it like this? - $ \int_0^{\infty} bv (v e^{-av^{2}}) , dv $ – Foon Jun 11 '15 at 10:39
1 Answers
0
First change variables so you're carrying less symbols around: set $x=v\sqrt{m/(2kT)}$, so the integral becomes $$ 4\pi \left(\frac{2kT}{m}\right)^{3/2} \int_0^{\infty} x^2 e^{-x^2} \, dx. $$ Now, the last can be done by parts: $$ \int_0^{\infty} 2x^2 e^{-x^2} \, dx = \left[ -xe^{-x^2} \right]_0^{\infty} + \int_0^{\infty} e^{-x^2} \, dx; $$ the first term vanishes and the second is well-known to be $\frac{1}{2}\sqrt{\pi}$, so the answer is $$ \int_0^{\infty} 4\pi v^2 e^{-mv^2/(2kT)} \, dv = \pi^{3/2}\left(\frac{2kT}{m}\right)^{3/2}. $$
(Remark: tabular integration is entirely equivalent to integration by parts, so provides nothing except neatness.)
-
Thanks Bro. That's a life saver. I didn't know that "well known" formula :) ... That's some awesome math right there!
(To your remark: I should've written substitution to tabular, my bad.)
– Foon Jun 11 '15 at 10:51 -
Edit: here's an extra 2 on the last line. It should be $$ \pi^{3/2}\left(\frac{2kT}{m}\right)^{3/2}. $$ – Foon Jun 11 '15 at 14:06
-