Integral of Bessel Functions Multiplying "polynomials"

Question

How can I compute the following integral: $$\int_{0}^{1} (1 - x^{2})^{\nu - \mu - 1} x^{\mu + 1} J_{\mu}(\alpha_{\nu}x) dx$$ where $\nu > \mu \geq 1$ and $J_{\nu }(\alpha_{\nu}) = 0$. The $J_{\nu}$ is the first kind Bessel function of order $\nu$.

Answer 1

If we use the expansion

$$J_{\mu}(\alpha_{\nu} x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \Gamma(k+\mu+1)} \left (\frac{\alpha_{\nu} x}{2} \right )^{\mu+2 k}$$

we end up getting, as the integral

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{k! \Gamma(k+\mu+1)} \left (\frac{\alpha_{\nu} }{2} \right )^{\mu+2 k} \int_0^1 dx \, (1-x^2)^{\nu-\mu-1} x^{2 \mu + 2 k+1} $$

The integral inside the sum is simply a beta integral and is equal to

$$\int_0^1 dx \, (1-x^2)^{\nu-\mu-1} x^{2 \mu + 2 k+1} = \frac12 \frac{\Gamma(\mu+k+1) \Gamma(\nu-\mu)}{\Gamma(k+\nu+1)}$$

Thus the integral is

$$\frac12 \Gamma(\nu-\mu) \left (\frac{\alpha_{\nu} }{2} \right )^{\mu-\nu} J_{\nu}(\alpha_{\nu}) = 0$$