Please could somebody tell me if there is a simpler form or known function to express the hypergeometric function next:
$_2F_3\left(\frac{1}{2},\frac{1}{2};1,1,\frac{3}{2};-4 \pi ^2 a^2\right)$
I am newby with that kind of functions.
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I do not think that this could simplify to anything. – Claude Leibovici Apr 14 '16 at 13:03
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A simple form (without infinite series) is an integral :
We start from the known integral of the squared Bessel function of the first kind :
$$\int_0^x \left(J_0(t)\right)^2 dt=x\:\:_2F_3\left(\frac{1}{2}\:,\:\frac{1}{2}\:;\:1\:,\:1\:,\:\frac{3}{2}\:;\:-x^2\right)$$ With $x=2\pi a$ : $$_2F_3\left(\frac{1}{2}\:,\:\frac{1}{2}\:;\:1\:,\:1\:,\:\frac{3}{2}\:;\:-4\pi^2a^2\right) = \frac{1}{2\pi a}\int_0^{2\pi a} \left(J_0(t)\right)^2 dt$$
I don't think that a simpler closed form exists with standard special functions of lower level.
JJacquelin
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Thanks, Jacquelin and Claude... That hypergeometric function was obtained after integrate the squared Bessel function, like Jacquelin shows. – Daniel Apr 15 '16 at 14:57