Bessel's function

Question

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.

Answer 1

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.

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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.