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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 $n$?

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  • $\begingroup$ G.N. Watson, A Treatise on the Theory of Bessel Functions, ( 2nd.ed.), Cambridge University Press, 1966, page 32, gives Neumann's Series for $J^2_n(z)$ archive.org/details/ATreatiseOnTheTheoryOfBesselFunctions/page/… $\endgroup$ – James Arathoon Jan 29 at 21:22
  • $\begingroup$ @JamesArathoon the series on page 32 looks different from the one you wrote in your answer. Is there an equivalence? $\endgroup$ – Alessandro Zunino Jan 29 at 21:57
  • $\begingroup$ Have you considered changing the starting point of the sum to $n$? From Neumann's formula I immediately get $$J^2_n(x)=\sum _{k=n}^{\infty } \frac{ (-1)^{k-n} (2 k)! }{2^{2 k} (k!)^2 (k+n)! (k-n)!}x^{2 k}$$ $\endgroup$ – James Arathoon Jan 29 at 22:27
  • $\begingroup$ Alternatively you can adjust the integral at the bottom of my answer to include $n$. $\endgroup$ – James Arathoon Jan 29 at 22:45
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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 $a$:

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

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  • $\begingroup$ I would be much more interested in a more approximated solution, but more readable. $\endgroup$ – Alessandro Zunino Jan 28 at 19:46
  • $\begingroup$ What $T$ stands for in this expression? $\endgroup$ – Alessandro Zunino Jan 28 at 21:06
  • $\begingroup$ The same $T$ as you have in the question. $\endgroup$ – Robert Israel Jan 28 at 23:43
  • $\begingroup$ Sorry, I have mistaken the subscript and interpreted it as a function. $\endgroup$ – Alessandro Zunino Jan 29 at 1:12
  • $\begingroup$ I may be doing something wrong, but I tried to plot the function (using $n$ as the independent variable) you wrote and the one I obtained using numerical integration and they look very different (your function looks roughly like a gaussian and the numerical result roughly like a sigmoid). $\endgroup$ – Alessandro Zunino Jan 29 at 13:59
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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).

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  • $\begingroup$ How did you get this solution? $\endgroup$ – Alessandro Zunino Jan 29 at 13:56
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    $\begingroup$ Hopefully this answers the question in the above comment and helps lead you to a proof. $\endgroup$ – James Arathoon Jan 29 at 14:29
  • $\begingroup$ Can you please provide a reference or a way to calculate the series of $J_n^2(x)$? $\endgroup$ – Alessandro Zunino Jan 29 at 19:44
  • $\begingroup$ I used Mathematica to calculate the square of the series starting with $J_0$. I suppose you could write out a Cauchy product to calculate the square of the series by hand. I found a pattern for $J_n$ by investigating the first few orders in Mathematica and making use of OEIS. Therefore I didn't develop a proof for the squaring of a Bessel function of arbitrary order $n$. $\endgroup$ – James Arathoon Jan 29 at 21:40

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