Integral of a bessel function composed with a trigonometric function

Question

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 $n$?

Answer 1

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 $a$:

$$ \frac{1}{\pi} \sum_{k=0}^\infty \frac{(-1)^k \Gamma(n+k+1/2)^2}{(k+n)!^2(k+2n)!k!} a^{2k+2n}$$

Answer 2

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).