Harmonic oscillator phase space integral

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SUMMARY

The integral discussed is derived from Quantum Physics by Le Bellac, specifically the expression \(\int dxdp\;\delta\left( E - \frac{p^2}{2m} - \frac{1}{2}m\omega^2x^2 \right) f(E)\), which evaluates to \(\frac{2\pi}{\omega} f(E)\). The integral represents a line integral around the perimeter of an ellipse in phase space. The discussion highlights two approaches: parameterizing the ellipse and using the arc length segment, both of which lead to complications. A key property of the Dirac Delta distribution is also noted, which aids in solving such integrals.

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Evertje
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Hi all,

I am having trouble with a certain integral, which I got from Quantum Physics by Le Bellac:
\int dxdp\;\delta\left( E - \frac{p^2}{2m} - \frac{1}{2}m\omega^2x^2 \right) f(E)
The answer to this integral should be 2\pi / \omega\; f(E).

My attempts so far:
This integral is basically a line integral around the perimeter of the (in general) ellipse traced out in phase space. Therefore I tried parameterizing the ellipse using x = acos(t) and p = bsin(t). From here, I've tried two different things:
1) Even though the original integral is 2D, I rewrite both dx and dy to dt, and then finally arrive at (since now the argument of the dirac delta is always equal to 0, along the curve):
-ab \int_0^{2\pi} sin(t)cos(t)dt = 0
2) Rewrite using the arc length segment
ds = \sqrt{ x'^2 + p'^2 } dt
this results in a complicated integral, which I am unable to solve. Also, by trying to simply use approximations for the perimeter of an ellipse turns out in other answers :(.
 
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There's a property of the Dirac Delta distribution that is worth remembering, and I think will help you greatly with this integral:

\delta(f(x))=\sum_i \frac{\delta(x-x_i)}{|f'(x_i)|}

Where f(x) is any continuously differentiable function, with roots (assumed to be simple) x_i.
 
Thanks a lot gabbagabbahey! I have solved the problem :)!
 

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