Change of variables in a simple integral

[MathematicalPhysicist]

Homework Statement

I have the integral $\int_0^\infty dv v^2 x \frac{\partial g}{\partial x}$.
Using $x=\beta(\frac{1}{2}mv^2-\mu)$ we get:
$\int_{-\infty}^\infty dx x (x+\beta \mu)^{3/2}\frac{\partial g}{\partial x}$.

The last integral upto some constants which don't get integrated.

Relevant Equations

they appear already in the HW statement.

So we have $x=\beta(1/2 mv^2-\mu)$, i.e $\sqrt{2(x/\beta+\mu)/m}=v$.
$dv= \sqrt{2/m}dx/\sqrt{2(x/\beta+\mu)/m}$.

So should I get in the second integral $(x+\beta \mu)^{1/2}$, since we have: $$v^2 dv = (2(x/\beta+\mu)/m)\sqrt{2/m} dx/\sqrt{2(x/\beta+\mu)/m}$$

So shouldn't it be a power of 1/2 and not 3/2?

Thanks in advance!

 

[anuttarasammyak]

You show no explanation about $\frac{\partial g}{\partial x}$.

 

[MathematicalPhysicist]

You show no explanation about $\frac{\partial g}{\partial x}$.

Well, it's written that $x$ and $g$ both depend on $v^2$ only.
How does that fact change any thing here?

 

[MathematicalPhysicist]

Ah, besides the fact that $\partial g/\partial x = \partial g /\partial v \partial v / \partial x$, but I am not sure he changes it here that way...

 

[MathematicalPhysicist]

I believe it should be 1/2 and not 3/2 because in the SM of Reif on equation (8) of page 108 he changes the LHS to 1/2 instead of 3/2.

Reading the SM alongside the problems and trying to understand the solutions can be quite time consuming, but hey that's what I chose to do. :-)

 

[anuttarasammyak]

How about checking it by looking at dimension ? Apart from $\frac{\partial g}{\partial x}$ integral has dimension of $L^4T^{-3}$ if it is correct to take v and x velocity and coordinate.