Sum of squared Fresnel sine integral I'm trying to find the following sum:
$$ \sum_{n=0}^{\infty} \frac{S\left(\sqrt{2n}\right)^2}{n^3}$$
where $S(n)$ is the fresnel sine integral, however, I think I made a mistake somewhere.
To start, I considered using parseval's identity:
$$ 2\pi\sum_{n=-\infty}^{\infty} |c_n|^2 = \int_{-\pi}^{\pi} |f(x)|^2 \space dx$$
where $f(x)$ is:
$$ f(x) = \sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} $$
$c_n$ becomes:
$$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} \right ) e^{-inx}\space dx $$
This integral is complicated, so I plugged it into wolfram alpha and found that
$$ c_n  = \frac{1}{2\pi} \left(-\sqrt{2\pi} \cdot \frac{S\left(\sqrt{2n}\right)}{n^{3/2}} \right)$$
so,
$$ |c_n|^2 = \frac{1}{4\pi^2} \left(2\pi \cdot \frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \frac{1}{2\pi} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right)$$
I think $|c_n|^2$ is finite for all n and is an even function of n. If this is true, then parseval's identity gives:
$$ 2\pi\sum_{n=-\infty}^{\infty} \frac{1}{2\pi} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx$$
and if $|c_n|^2$ is even then this expression becomes:
$$ 2\sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx$$
I believe that
$$ \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx = \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right)^2 \space dx$$
and if I plug in the second integral into wolfram alpha again, I find that (EDIT user Claude Leibovici correctly found that):
$$ \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right)^2 \space dx = \pi^2$$
So, in total I have:
$$ 2\sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \pi^2$$
or 
$$ \sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \frac{\pi^2}{2}$$
The problem is that wolfram alpha suggests that the sum approaches .549, but my answer is ~4.93. Where did I make a mistake?
 A: With help from reddit user GamblingTheory the solution is as follows:
let
$$ f(x) = -\sqrt{2\pi} * \left( \sqrt{\frac{ix}{2}} + \sqrt{-\frac{ix}{2}} - \sqrt{\pi}\right)$$
thus,
$$ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) * e^{-inx} \space dx = \frac{S({\sqrt{2n}})}{n^{3/2}}$$
Using Parseval's Identity:
$$\sum_{-\infty}^{\infty}|c_n|^2 = \sum_{-\infty}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} -\sqrt{\pi}\right|^2 \space dx = \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} - \sqrt{\pi}\right)^2 \space dx = \frac{\pi^{2}}{3}$$
We rewrite the sum like so:
$$\sum_{-\infty}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} = 2\sum_{1}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} + \frac{2\pi^2}{9}$$
thus
$$2\sum_{1}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} + \frac{2\pi^2}{9} = \frac{\pi^{2}}{3}$$
$$\sum_{1}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} = \frac{\pi^{2}}{6} - \frac{\pi^2}{9} = \frac{\pi^2}{18}$$
I'm not sure if this sum is useful in anyway, but I thought it was a fun sum to compute especially because it is reminiscent of:
$$ \sum_{1}^{\infty} \frac{1}{n^{3}}  $$
A: This is not answer.
I am stuck with the problem but I have a few remarks


*

*I suppose that the summation starts at $n=1$ and not at $n=0$

*Using Wolfran Alpha (see here)
$$\int_{-\pi}^\pi\left(\frac{\sqrt{-i x}}{\sqrt{2}}+\frac{\sqrt{i x}}{\sqrt{2}}\right)^2\,dx= \pi^2$$

*Using Wolfran Alpha (see here)
$$\int_{-\pi}^\pi\left|\frac{\sqrt{-i x}}{\sqrt{2}}+\frac{\sqrt{i x}}{\sqrt{2}}\right|^2 \,dx= \pi^2$$
