<s>[INST] Hypergeometric representation of Fresnel $S(x)$ 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.
 [/INST] 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 $S_0 = \frac{1}{3}$.
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