Integral of a bessel function composed with a trigonometric function Is there any standard way or approximated way to calculate an integral of the form
$$
\int_0^T J_n^2[a\cos(2\pi t/T)]dt
$$
where $J_n$ is the bessel function of first kind of order $n$?
 A: Using Maple I get a hypergeometric expression:
$$ {\frac {T{\mbox{$_2$F$_3$}(1/2+n,1/2+n;\,n+1,n+1,1+2\,n;\,-{a}^{2})}
\Gamma \left( 1/2+n \right) {a}^{2\,n}{4}^{-n}}{\sqrt {\pi} \left( n!
 \right) ^{3}}}
$$
EDIT: This is really telling you the series expansion in powers of $a$:
$$ \frac{1}{\pi} \sum_{k=0}^\infty \frac{(-1)^k \Gamma(n+k+1/2)^2}{(k+n)!^2(k+2n)!k!} a^{2k+2n}$$
A: This is another interesting conjectural result for any order of Bessel function of the first kind which can be compared with Robert Israel's Maple result. 
$$\int_0^T J_n^2[a\cos(2\pi t/T)]\,dt=T \,\sum _{k=0}^{\infty } \left(\frac{(-1)^{k+n} \binom{2 k}{k}^2  a^{2 k} } {(k!)^2\, 4^{2 k}   } \frac{\prod _{j=0}^{n-1} (k-j)}{\prod _{j=0}^{n-1} (j+k+1)}\right)$$
This result came from the inspection of series of $[J_n(x)]^2$
$$[J_n(x)]^2=\sum _{k=0}^{\infty } \left(\frac{(-1)^{k+n} \binom{2 k}{k}   }{ (k!)^2  \,2^{2 k}}\frac{\prod _{j=0}^{n-1} (k-j)}{\prod _{j=0}^{n-1} (j+k+1)} x^{2 k}\right)\tag{1}$$
together with the evaluation of the integral
$$\int_0^T \left(a \cos \left(\frac{2 \pi  t}{T}\right)\right)^{2 k} \, dt=T\;\frac{   \Gamma \left(k+\frac{1}{2}\right)}{\sqrt{\pi }\; \Gamma (k+1)}a^{2 k}=x^{2k}$$
So the conjectural step is equation (1).
