Integration by parts

Question

Hi,

I am studying a physics text and I came across a derivation that I cannot reproduce myself.
It involves the INT[(v^2.exp(-alpha.v^2).dv] with integration limits between zero and infinity.
This is one of the socalled Gaussian integrals.Here alpha is a constant and v is a velocity.
In the text(Principles of Modern Physics by A.P.French),this integral is initially treated by
integration by parts

They say that it is equal to:[-(v.exp(-alpha. v^2)/2.alpha] + ( 1/(2.alpha).INT[(exp(-alpha.v^2).dv] ( Integral between zero and infinity again)
When I do this ,I agree with this calculation,except that I find that in the last integral it should be :(1/(alpha).INT[(exp(-alpha.v^2).dv]
Which is twice the value obtained in the text.

My reasoning: integration by parts: d(a.b)= a.db +b.da
Here :a=(v^2) and da=2.v.dv

b=-[(1/(2.alpha.v).exp(-alpha.v^2)] and db=exp(-alpha.v^2)

In this case the integral in question becomes INT[b.da] or

INT[ -[(1/(2.alpha.v).exp(-alpha.v^2)]. 2.v.dv] which simplifies to

(1/alpha). INT[(exp(-alpha.v^2).dv] and NOT ( 1/(2.alpha).INT[(exp(-alpha.v^2).dv].

Where am I making a mistake here?
Thank you for any input!


Eric T. S.

Answer 1

∫₀^∞v²e^{-αv^2}dv

If a=v², then da=2vdv, That's correct. Next, the remaing under the integral is d(b)=e^{-αv^2}dv, from which you cannot find b. Therefore, you take as a just v (a=v) and db=ve^{-αv^2}dv, from where you can find b easily.

db=ve^{-αv^2}dv= -(1/2α)e^{-αv^2}d(-αv²)=-(1/2α)d(e^{-αv^2})

now you can see that b=-(1/2α)e^{-αv^2}.

Now if you integrate by parts, you will have

=-v(1/2α)e^{-αv^2}+∫e^{-αv^2}dv.

 

Integration by parts:∫f(x)g'(x)dx=f(x)g(x)-∫g(x)f'(x)dx.