I am given the potential:

$$\phi (x, y) = \frac{k}{2}(x^2+y^2) + axy$$

Where a is a constant.

I have to compute:

$$\int_{-\infty}^\infty \int_{-\infty}^\infty \mathrm e^{-\beta\phi}dxdy$$

Changing to polar coordinates:

$$\int_{0}^{2\pi}\int_{0}^{b} e^{-\beta\phi}rdrd\theta$$

What I have done is:

$$\iint e^{-\beta\phi}rdrd\theta = \iint e^{-\frac{\beta Kr^2}{2}} e^{-\beta a r^2 cos\theta sin\theta}rdrd\theta = \iint e^{-\beta f(r, \theta)}rdrd\theta$$


$$f(r, \theta) = \frac{Kr^2}{2} [(1+\frac{cos \theta sin \theta}{K})^2 -(\frac{cos \theta sin \theta}{K})^2]$$

Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.


Using polar coordinates here is not the best method.

Take squares and try:

$$x = u + v$$

$$y = u - v$$

  • 1
    $\begingroup$ Is the integral meant to be indefinite? $\endgroup$ – J.G. Nov 22 '18 at 11:55
  • $\begingroup$ @J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point. $\endgroup$ – JD_PM Nov 22 '18 at 12:04
  • $\begingroup$ See math.stackexchange.com/questions/1096793/… $\endgroup$ – JD_PM Nov 23 '18 at 15:34

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