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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?
Thanks in advance!
##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!