I am trying to prove that
$ J_n^2(z) = \frac{2}{\pi}\int_0^\frac{\pi}{2}{J_{2n}(2z\cos(\theta))d\theta}, $
and for some reason I am failing to prove it. I am using the integral representation and shuffling with the cosines but going nowhere. Any tips?
Let us try the brute-force approach. We have: $$ J_n(x)=\sum_{a\geq 0}\frac{(-1)^a}{a!(a+n)!}\left(\frac{x}{2}\right)^{2a+n} \tag{1}$$ hence: $$ J_n^2(x) = \sum_{m\geq 0}\sum_{a+b=m}\frac{(-1)^m (x/2)^{2m+2n}}{a!b!(a+n)!(b+n)!} \tag{2}$$ where: $$\begin{eqnarray*} \sum_{a+b=m}\frac{1}{a!b!(a+n)!(b+n)!}&=&\frac{1}{m!(m+2n)!}\sum_{a+b=m}\binom{m}{a}\binom{m+2n}{b+n}\\&=&\frac{1}{m!(m+2n)!}[x^{m+n}](1+x)^m (1+x)^{m+2n}\\&=&\frac{1}{m!(m+2n)!}\binom{2m+2n}{m+n}\tag{3}\end{eqnarray*}$$ leads to: $$ J_n^2(x)=\sum_{m\geq 0}\frac{(-1)^m (x/2)^{2m+2n}}{m!(m+2n)!}\binom{2m+2n}{m+n}.\tag{4} $$ Since $\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2h}(\theta)\,d\theta = \frac{1}{4^h}\binom{2h}{h}$ follows from De Moivre's formula, in order to prove the claim it is enough to expand $J_{2n}(2z\cos\theta)$ as a power series in $2z\cos\theta$, perform terwmwise integration and exploit $(4)$. The claim is ultimately a consequence of Vandermonde's identity proved in $(3)$.