Hypergeometric representation of Fresnel $S(x)$

Question

I am trying to find a representation for the Fresnel integral $$S(x)=\int_0^x\sin\frac{\pi t^2}{2}\,\mathrm dt$$ Then with $$\sin x=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ We have $$S(x)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}\int_0^x\left(\frac{\pi t^2}2\right)^{2n+1}\mathrm dt$$ $$S(x)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}\left(\frac\pi2\right)^{2n+1}\frac{x^{4n+3}}{4n+3}$$ $$S(x)=\frac{\pi x^3}{2}\sum_{n\geq0}\frac{1}{(2n+1)!(4n+3)}\left[-\frac{\pi^2x^4}{4}\right]^{n}$$ Then setting $$S_n=\frac{1}{(2n+1)!(4n+3)}\left[-\frac{\pi^2x^4}{4}\right]^{n}$$ We have $$\frac{S_{n+1}}{S_n}=\frac{-\pi^2 x^4}{16(n+1)}\frac{n+3/4}{(n+7/4)(n+3/2)}$$ Hence we have $$S(x)=\frac{\pi x^3}2\,_1F_2\left(\frac34;\frac32,\frac74;-\frac{\pi^2x^4}{16}\right)$$ But the Wolfram Functions site Says that $$S(x)=\frac{\pi x^3}{\color{red}6}\,_1F_2\left(\frac34;\frac32,\frac74;-\frac{\pi^2x^4}{16}\right)$$ Where did that extra $1/3$ come from? Thanks.

Answer 1

Note that the recurrence relation for $(S_n)_{n \in \mathbb{N}_0}$ implies $$ S_n = \frac{\Gamma\left(n + \frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}\right) }\frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(n + \frac{3}{2}\right)} \frac{\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(n + \frac{7}{4}\right)} \frac{\left(-\frac{\pi^2 x^4}{16}\right)^n}{n!} \color{red}{S_0}$$ and we have $S_0 = \frac{1}{3}$.