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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.

  • 2
    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.2011-11-11
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    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 $\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty \phi(x^2+y^2)\delta'(x^2+y^2-a)\delta\left(x^2-y^2\right)  \; dx\;dy$.2011-11-11

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