How Is the Number of Molecules Calculated in Maxwell Speed Distribution?

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Homework Statement


Show that the number N(0,v), of molecules of an ideal gas with speeds between 0 and v is given by

N(0,v) = N \left[ erf(\xi) - \frac{2}{\sqrt{\pi}} \xi e^{-\xi^2} \right]

Where,

erf(\xi) = \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{-x^2} dx

And,

\xi^2 = \left(\frac{mv^2}{2kT} \right)

Homework Equations



\frac{dN_v}{N} = \sqrt{\frac{2}{\pi}} \left( \frac{m}{kT}\right)^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2kT}} \ dv

The Attempt at a Solution



Alright, so I managed to get to the following

\frac{dN_v}{N} = \frac{4}{\sqrt{\pi}} x^2 e^{-x^2} dx

Where,

\alpha = \frac{m}{2kT}, \ x = \sqrt{\alpha}v

=========================\\===========================

So far so good, but now when checking the solution on the textbook it claims the following algebraic manipulation which I can't follow

\frac{2N}{\sqrt{\pi}}\int_{0}^{\xi} x (2xe^{-x^2} dx) = \frac{2N}{\sqrt{\pi}} x e^{-x^2} |_{\xi}^{0} \ + \ N \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{-x^2} dx

What was done on the integral above, how can you "split" it like that?
 
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Integration by parts
 


It's integration by parts. udv=d(uv)-vdu with u=x and dv=(2x)e^(-x^2)dx.
 
L-x said:
Integration by parts

But of course! Agh, how didn't I see that, I feel very stupid right now. :biggrin:

Thanks for the highlight, L-x!
 
No worries. It's not immidiately obvious, although the integral is written in a way which gives you a clue: x(2x) instead of 2x^2
 
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