Abel transform to Sum over sin 2 ( n x )

MAIN If we have ^\infty {n} try introducing a parameter r → 1, this gives ^\infty {n} then try an Abel transform with respect to r, knowing that if r = 1 then r² = 1, K = 2\int_y^\infty {} \; dr = }{4}{n(n^2 x^2)^{1/4}} Then sum over this ^\infty K = {8}-{8})}{(x^2)^{1/4}} x y(_{3/2}(e^{-2ixy^2})+i _{3/2}(e^{2ixy^2})) by the inverse transform we should then have ^\infty {n} = -{2}+{2})}{} \int_r^\infty _{1/2}(e^{-2ixy^2})-i_{1/2}(e^{2ixy^2}))}{(x^2)^{1/4}} \; dy ^\infty {n} = -{2}+{2})}{} \int_1^\infty _{1/2}(e^{-2ixy^2})-i_{1/2}(e^{2ixy^2}))}{(x^2)^{1/4}} \; dy