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I read that $\int\limits_0^1 xJ_n(j_{na}x) J_n(j_{nb}x) dx={1\over 2}\delta_{ab}[J_n'(j_{na})]^2$, where $j_{na},j_{nb}$ are zeros of $J_n$, the Bessel function of the $n$th degree. Is there a simple proof for this? Thanks.

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Where did you read this ? –  Sasha Nov 4 '11 at 14:54
@Sasha: On quite a few websites talking about Bessel functions. This demonstrates the orthogonality and helps us normalize... But I have't found a website that gives a proof of this. They tend to say "it can be shown that..." I have tried proving it myself, starting with the Bessel equations. I have managed to show that if $a\neq b$ tgen tge integral vanishes, but I can't show the $a=b$ case. Thought I might let $a=b+\epsilon$ Then take limit as epsilon tends to 0... –  Gordon Nov 4 '11 at 15:14
I'm tempted to think this can be proven as a special case of the orthogonality result in Sturm-Liouville theory: [url]en.wikipedia.org/wiki/Sturm–Liouville_theory[\url]. I'll leave the details to someone else though. –  Ragib Zaman Nov 4 '11 at 15:22
@Ragib I suspect that the reason why the URL you posted isn't working is that there's an en-dash in the title. I also suspect that a Wikipedia page with the same title but a hyphen rather than an en-dash is a redirect page that directs to the article whose title has an en-dash. Let's try it: [url]en.wikipedia.org/wiki/Sturm-Liouville_theory[/url] –  Michael Hardy Nov 4 '11 at 16:31
Typo. I'll try again: [url]en.wikipedia.org/wiki/Sturm-Liouville_theory[\url] –  Michael Hardy Nov 4 '11 at 16:32
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1 Answer

Using Mellin convolution technique, the following integral is available in closed form, assuming $n \in \mathbb{Z}^+$: $$ \int_0^1 x J_n\left( \alpha x \right) J_n \left( \beta x \right) \mathrm{d} x = \frac{\beta J_n(\alpha ) J_{n-1}(\beta )-\alpha J_{n-1}(\alpha ) J_n(\beta)}{ \alpha^2-\beta ^2} $$ From this, it follows immediately that if $\alpha = j_{na}$ and $\beta = j_{nb}$ and $a \not= b$, the integral vanishes.

Taking the limit of $\beta \to \alpha$: $$ \int_0^1 x J_n\left( \alpha x \right) J_n \left( \alpha x \right) \mathrm{d} x = \frac{1}{2} \left( J_n^2(\alpha) - J_{n+1}(\alpha) J_{n-1}(\alpha) \right) $$ When $\alpha = j_{na}$ the first term vanishes. Using the fact that: $$ J_n^\prime(x) = \frac{n}{x} J_n(x) - J_{n+1}(x) = J_{n-1}(x) - \frac{n}{x} J_n(x) $$ and setting $x = j_{na}$ we recover the identity you encountered.

Actually, I am wrong about need to use Mellin convolution technique. The integrand has a closed-form anti-derivative: $$ \int x J_n\left( \alpha x \right) J_n \left( \beta x \right) \mathrm{d} x = \frac{\beta x J_n(x \alpha ) J_{n-1}(x \beta )-\alpha x J_{n-1}(x \alpha ) J_n(x \beta )}{\alpha ^2-\beta ^2} $$ which is easy to check by differentiation.

After that the definition integral is obtained using the fundamental theorem of calculus.

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Thanks, Sasha! Is there an analytic way to show the 1st integral? –  Gordon Nov 4 '11 at 15:59
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