<s>[INST] How to derive this Bessel function relationship? $$J_n(x) = \frac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt$$
 [/INST] 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}
 $
</s>