# Change of variables in a simple integral

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## 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?

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anuttarasammyak
Gold Member
You show no explanation about ##\frac{\partial g}{\partial x}##.

Gold Member
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?

Gold Member
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...

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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
Gold Member
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.