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There exists the following relation for the Besselfunctions $j_{l}(\alpha\,x)$, $j_{l}(\beta\,x)$ with $\alpha$, $\beta$ $\in \mathbb{R}$ and $x\in \mathbb{R}$

$$\int_{0}^{\infty}dx \, x^{2}\, j_{l}(\alpha\,x)\, j_{l}(\beta\,x) = \frac{\pi}{\beta}\, \delta(\alpha^{2}-\beta^{2}). $$

Now, assume $\alpha_{n} , \beta_{n} \in \mathbb{Z}$ , i.e. are discrete numbers, then the above equation can be rewritten in terms of

$$\int_{0}^{\infty}dx \, x^{2}\, j_{l}(\alpha_{n}\,x)\, j_{l}(\beta_{n}\,x) = \int_{0}^{\infty} \,dy \, dz \, \delta(y-\alpha_{n}) \, \delta(z-\beta_{n}) dx \, x^{2}\, j_{l}(y\,x)\, j_{l}(z\,x)$$ $$= \int_{0}^{\infty} \,dy \, dz \, \delta(y-\alpha_{n}) \, \delta(z-\beta_{n}) \frac{\pi}{z}\, \delta(y^{2}-z^{2})$$ $$=\frac{\pi}{\beta_{n}}\, \delta(\alpha_{n}^{2}-\beta_{n}^{2}). $$

Question: I don't understand how the Delta Distribution which is supposed to be continuous can be understood in terms of discrete variables, since the Dirac Delta within the theory of generalized functions is always used with some test function which is integrated over which is not possible in this case.

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Ah, physicist notation, how you always find a way to confuse me. Anyway OP this looks like you just took a Dirac delta function, arbitrarily restricted its arguments to integers and falsely assumed it still lives inside a distributional framework afterwards. At any rate it still formally expresses when the integral diverges and what its value is when it converges. – anon Sep 13 '11 at 8:11
You could check this. – Did Oct 19 '11 at 9:18

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