# Integration error spotting

What have I done wrong?

I have to evaluate the following integral:

$$\int\limits_0^\infty\int\limits_0^{2\pi} \phi(r^2)\delta'(r^2-a)\delta\left(\theta- \left(n+{1\over 2}\right){\pi\over 2}\right) r \; d\theta \;dr$$

My (wrong) working:

$\implies 4\int\limits_0^\infty \phi(x)\delta'(x-a){1\over 2}dx$ by letting $x=r^2$

$\implies -2\phi'(a)$

I should be getting the integral equalling $\phi(a)-\phi'(a)$.

Thank you.

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Where does the $4$ come from? And what happened to the integration over $\theta$? It should give $0$ for most $n$. Finally, I doubt that the result can be $\phi(a)-\phi'(a)$, since those two terms have different units and there's nothing in the integrand that could cause that mismatch. – joriki Nov 11 '11 at 22:50
Thanks, @joriki. Isn't $\int\limits_0^{2\pi} \delta\left(\theta- \left(n+{1\over 2}\right){\pi\over 2}\right) d\theta =4$? Since there are 4 values of $\theta\in [0,2\pi)$ s.t. we have $\left(n+{1\over 2}\right){\pi\over 2}$? Actually this question has gone through a few steps beforehand, maybe I made a mistake there. The integral started life as $\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty \phi(x^2+y^2)\delta'(x^2+y^2-a)\delta\left(x^2-y^2\right) \; dx\;dy$. – sgt pepper Nov 11 '11 at 23:34

In the present case you made two mistakes in deriving what you at first presented as the given problem. One seems to be just a problem of notation – it seems that you intended this to be summed over all integer $n$, but that's not what you wrote. The other is that you dropped a factor of $r^2$ in the argument of the second $\delta$. Here's a calculation of your original integral:
$$\begin{eqnarray} \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty \phi(x^2+y^2)\delta'(x^2+y^2-a)\delta(x^2-y^2)\mathrm dx\mathrm dy &=& \int\limits_0^\infty\int\limits_0^{2\pi} \phi(r^2)\delta'(r^2-a)\delta(r^2\cos2\theta)\mathrm rd\theta\mathrm dr \\ &=& \frac12\int\limits_0^\infty\int\limits_0^{2\pi} \phi(u)\delta'(u-a)\delta(u\cos2\theta)\mathrm d\theta\mathrm du \\ &=& 2\int\limits_0^\infty\frac{\phi(u)}u\delta'(u-a)\mathrm du \\ &=& 2\int\limits_0^\infty\frac{\phi(u)-u\phi'(u)}{u^2}\delta(u-a)\mathrm du \\ &=& 2\frac{\phi(a)-a\phi'(a)}{a^2} \;. \end{eqnarray}$$
This is also not the result $\phi(a)-\phi'(a)$ that you expected. Perhaps you're assuming $a=1$ somewhere? Then there would only be an excess factor of $2$.