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How to do this integration?

$$ \int_{-\infty}^{\infty}dq\int_{-\infty}^{\infty} dp \; \delta(E-\frac{p^2}{2m}-\frac{k}{2}q^2)= 2\pi\sqrt{\frac{m}{k}}$$

I obtained the result using Mathematica, I am not even sure it is correct. Anyways, I'd love to know how one can evaluate this by hand. I am familiar with the Delta function and the identities on how to evaluate those integrals over infinities.

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  • $\begingroup$ It's not any different from doing one integral at a time. Just ignore the $dq$ integral and treat $q$ as a constant while doing the $dp$ integral. $\endgroup$ – knzhou May 13 '17 at 22:49
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic May 13 '17 at 23:17
  • $\begingroup$ I firmly believe that integrals arising in statistical physics like this one should be posted here. I've often been told on Mathematics to ask physicists because mathematicians don't do integrals. $\endgroup$ – Bob Knighton May 14 '17 at 10:40
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The simplest way to solve this - and particularly, the way that minimizes the chances of messing it up - is to switch over to a single coordinate that inside the delta function. In your case it's easy - just choose an appropriate polar representation: set \begin{align} q& = A\, r\cos(\theta)\\ p& = B\, r\sin(\theta), \end{align} and require that $\frac{1}{2m}A^2 = \frac{k}{2}B^2$ (via e.g. $A=1$, $B=1/\sqrt{mk}$) to get \begin{align} \int_{-\infty}^\infty \mathrm dp\int_{-\infty}^\infty \mathrm dq \,\delta(E-\frac{1}{2m}p^2-\frac{k}{2}q^2) &= B\int_0^\infty r\:\mathrm dr \int_0^{2\pi}\mathrm d\theta \, \delta(E-\frac{1}{2m}r^2) \\&= \frac{2\pi}{\sqrt{mk}}\int_0^\infty \delta(E-\frac{1}{2m}r^2) r\:\mathrm dr. \end{align} From there, change variables to $u=r^2/2m$, so that $\mathrm du=r\,\mathrm dr/m$, which gives you \begin{align} \int_{-\infty}^\infty \mathrm dp\int_{-\infty}^\infty \mathrm dq \,\delta(E-\frac{1}{2m}p^2-\frac{k}{2}q^2) &= \frac{2m\pi}{\sqrt{mk}}\int_0^\infty \delta(E-u) \mathrm du, \end{align} and that reduces to the result you quote since $\int_0^\infty \delta(E-u) \mathrm du=1$ whenever $E>0$.

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  • $\begingroup$ The way you chose the substitution, I actually got $\frac{k}{2} A^2 = \frac{B^2}{2m}$ and therefor $A=1$ and $B=\sqrt{km}$. Anyways, I was able to follow your steps and will definitely add this substitution to my tool box. One last question, do we need $E>0$ or is it $E\geq 0$ still sufficient for the last integral to give 1? $\endgroup$ – Marsl May 15 '17 at 18:55
  • $\begingroup$ @Marsl The delta function at the edges of domains isn't particularly well defined - but it's only one point and it can be set at will without changing the results of whatever further integration you're doing over $E$ later. If you get to a point where that choice matters, you've probably done something wrong. $\endgroup$ – Emilio Pisanty May 15 '17 at 19:09
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The idea here is to use the following property of the delta function:

$$\int_{-\infty}^{\infty}\delta(f(x))g(x)\,\mathrm{d}x=\sum_{r}\frac{g(r)}{|f'(r)|}$$

Where the sum ranges over all values of $r$ such that $f(r)=0$. This is basically just what happens when you change variables to perform the integral.

If we let $f(p)=p^2/2m+kq^2/2-E$, then $f=0$ at $p_{\pm}=\pm\sqrt{2mE-kmq^2}$, which is real only for $kq^2\leq 2E$. We also have $|f'(p_{\pm})|=\sqrt{(2E-kq^2)/m}$, so long as $kq^2\leq 2E$. Thus, we can perform the $p$ integral to get

$$2\int_{-\sqrt{2E/k}}^{\sqrt{2E/k}}\mathrm{d}q\,\sqrt{\frac{m}{2E-kq^2}}=\sqrt{\frac{4m}{k}}\int_{-1}^{1}\frac{\mathrm{d}u}{\sqrt{1-u^2}}=2\pi\sqrt{\frac{m}{k}}$$

Which is exactly what you got!

I hope this helped!

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  • $\begingroup$ Hi, there, thanks for the help! I am not familiar with the formula you gave there but found it on wikipedia. Unfortunately, I still miss some points. 1.The 2 in front is because we have 2 terms contributing to the sum, namely $p_+$ and $p_-$? 2. I suppose that $g(r) = 1$ ? and 3. how did you get the new limits $\sqrt{2E/k}$ for the q-integral? $\endgroup$ – Marsl May 15 '17 at 19:06
  • $\begingroup$ You are exactly correct. The factor of $2$ is due to the fact that we have two terms contributing the same amount. In this case $g(r)=1$, since no function multiplies the delta function in the original integral. Finally, the bounds $\pm\sqrt{2E/k}$ come from the fact that the argument can only be set to zero if the value of $q$ is within those bounds. It's important to always be careful to check that a solution to the delta function exists, since that solution defines the domain of integration once you've done the first integral. (Also I assumed everyhwere that $E\geq 0$.) $\endgroup$ – Bob Knighton May 15 '17 at 22:49

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