How to prove that $J_\frac{-5}{2}(x)= \frac{\sqrt2}{\sqrt{x\pi}}[\frac{3}{x}\sin x+\frac{3-x^2}{x^2}\cos x]$

Question

How to prove that $$J_\frac{-5}{2}(x)= \sqrt{\frac{2}{\pi x}}\left(\frac{3}{x}\sin x+\frac{3-x^2}{x^2}\cos x\right)$$

I want to do this by using the definition of $J_{-n}(x)$ then putting value of $n=-5/2$. But the problem is that I do not know how to evaluate the gamma function involved in series such as $\Gamma(-3/2)$ , $\Gamma(-1/3)$, $\Gamma(1/3)$ etc

Thanks.

Answer 1

For $\Gamma$ of half-integer values, use $\Gamma(1/2) = \sqrt{\pi}$ together with the functional equation $\Gamma(z+1) = z \Gamma(z)$ to go either up or down by $1$.

added

Half-integer Gammas are the only ones you need in $J_{-5/2}$. You do not need $\Gamma(1/3)$ or $\Gamma(-1/3)$.

Answer 2

You do not have to evaluate any value of the $\Gamma$ function: you just have to check that both the RHS and the LHS satisfy the Bessel differential equation for $n=-\frac{5}{2}$, and that the RHS is unbounded in a right neighbourhood of zero.

You may also go through Frobenius method, by slightly changing lines from $(4)$ to $(26)$ here.