How to derive this Bessel function relationship?

Question

$$J_n(x) = \frac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt$$

Answer 1

Show that

$J_n(x) = \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt $

I did my usual naive plugging in of definitions and, with some help from Wolfy, it worked out.

A pleasant surprise.

Since

$J_a(x) =\sum_{m=0}^{\infty} \dfrac{(-1)^m}{m!(m+a)!}\dfrac{x^{2m+a}}{2^{2m+a}} $ and, according to Wolfy, $\int_0^1 (1 - x^2)^a x^b dx = \dfrac{Γ(a + 1) Γ((b + 1)/2)}{2 Γ(a + b/2 + 3/2)} $ for $Re(b)>-1 ∧ Re(a)>-1 $,

$\begin{array}\\ U_{n,m}(x) &=\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt\\ &=\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt\\ &=\int_0^1(1-t^2)^{n-m-1}t^{m+1}\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}t^{2k+m}}{2^{2k+m}}dt\\ &=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\int_0^1(1-t^2)^{n-m-1}t^{m+1}t^{2k+m}dt\\ &=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\int_0^1(1-t^2)^{n-m-1}t^{2k+2m+1}dt\\ &=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\dfrac{Γ(n-m) Γ((2k+2m+2)/2)}{2 Γ(n-m-1 + (2k+2m+1)/2 + 3/2)}\\ &=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\dfrac{Γ(n-m) Γ(k+m+1}{2 Γ(n-m-1 + k+m+2)}\\ &=\dfrac{x^{m}}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k}}{2^{2k}}\dfrac{Γ(n-m) Γ(k+m+1}{2 Γ(n+k+1)}\\ &=\dfrac{x^{m}\Gamma(n-m)}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k}}{2^{2k}}\dfrac{ Γ(k+m+1)}{2 Γ(n+k+1)}\\ &=\dfrac{x^{m}\Gamma(n-m)}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k}}{2^{2k}}\dfrac{ (k+m)!}{2(n+k)!}\\ &=\dfrac{x^{m}\Gamma(n-m)}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!2(n+k)!}\dfrac{x^{2k}}{2^{2k}}\\ &=\dfrac{\Gamma(n-m)}{2}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(n+k)!}\dfrac{x^{2k+m}}{2^{2k+m}}\\ &=\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(n+k)!}\dfrac{x^{2k+n}}{2^{2k+n}}\\ &=\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}J_n(x)\\ \end{array} $

so

$\begin{array}\\ \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}U_{n, m}(x) &=\dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt\\ &= \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}J_n(x)\\ &= \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}J_n(x)\\ &= J_n(x)\\ \end{array} $