Sum of squared Fresnel sine integral

Question

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?

Answer 1

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:

$$ \sum_{1}^{\infty} \frac{1}{n^{3}} $$

Answer 2

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) $$\int_{-\pi}^\pi\left|\frac{\sqrt{-i x}}{\sqrt{2}}+\frac{\sqrt{i x}}{\sqrt{2}}\right|^2 \,dx= \pi^2$$