Apery and Fresnel join forces

Parseval's theorem states that, for a Fourier series $\displaystyle f(x) = \frac{1}{2}a_0 + \sum_{n=1}^{\infty} a_n\cos(nx) + b_n\sin(nx)$ with

$\begin{aligned} a_0 &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \ dx \\ a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) \ dx \\ b_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) \ dx \end{aligned}$

we have

$\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)^2 \ dx = \frac{1}{2}a_0^2 + \sum_{n=1}^{\infty} a_n^2 + b_n^2$

Since we are considering $f(x) = |x|^{\sigma}$ , we get that

$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x|^{\sigma}\sin(nx) = 0$

because $|x|^{\sigma}\sin(nx)$ is an odd function. Thus, we only need to deal with the $\frac{1}{2}a_0$ and $a_n\cos(nx)$ components. We can guess that $a_n = k\dfrac{S(\sqrt{2n})}{n^{3/2}}$ , where $k$ is some constant, so we will find the value of $\sigma$ from that.

$\begin{aligned} a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} |x|^{\sigma}\cos(nx) \ dx \\ &= \frac{2}{\pi} \int_0^{\pi} x^{\sigma}\cos(nx) \ dx & (|x|^{\sigma}\cos(nx) \text{ is even}) \\ &= -\frac{2}{\pi} \int_0^{\pi} \sigma x^{\sigma-1} \cdot \frac{1}{n} \sin(nx) \ dx & (\text{Integration by parts}) \\ &= -\frac{2\sigma}{n^2} \int_0^{\sqrt{2n}} t \left( \frac{\pi}{2}t^2 \right)^{\sigma-1}\sin \left( \frac{\pi}{2}t^2 \right) \ dt & (\text{Let } nx = \frac{\pi}{2}t^2) \\ &= -\frac{2\sigma}{n^2} \left( \frac{\pi}{2n} \right)^{\sigma-1} \int_0^{\sqrt{2n}} t^{2\sigma-1} \sin \left( \frac{\pi}{2}t^2 \right) \ dt \end{aligned}$

We are very close to getting the Fresnel sine integral. If we let $\sigma = \frac{1}{2}$ to eliminate the $t^{2\sigma-1}$ in the integral, we get

$a_n = -\frac{2\sigma}{n^2} \left( \frac{\pi}{2n} \right)^{\sigma-1} \int_0^{\sqrt{2n}} t^{2\sigma-1} \sin \left( \frac{\pi}{2}t^2 \right) \ dt = -\sqrt{\frac{2}{\pi}}\frac{S(\sqrt{2n})}{n^{3/2}}$

Hence, $\sigma = \frac{1}{2}$ and $k = -\sqrt{\frac{2}{\pi}}$ . Now, we need to find the value of $a_0$ , which proves to be simple:

$\begin{aligned} a_0 &= \frac{1}{\pi} \int_{-\pi}^{\pi} |x|^{1/2} \ dx \\ &= \frac{2}{\pi} \int_0^{\pi} x^{1/2} \ dx & (|x|^{1/2} \text{ is even}) \\ &= \frac{2}{\pi} \frac{\pi^{3/2}}{3/2} \\ &= \frac{4}{3}\pi^{1/2} \end{aligned}$

So, using Parseval's theorem, we have that

$\begin{array}{ll} \displaystyle \frac{1}{2}a_0^2 + \sum_{n=1}^{\infty} a_n^2 &= \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)^2 \ dx \\ \displaystyle \frac{1}{2}\left( \frac{4}{3}\pi^{1/2} \right)^2 + \sum_{n=1}^{\infty} \left( -\sqrt{\frac{2}{\pi}} \frac{S(\sqrt{2n})}{n^{3/2}} \right)^2 &= \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} (|x|^{1/2})^2 \ dx \\ \displaystyle \frac{8}{9}\pi + \sum_{n=1}^{\infty} \frac{2}{\pi}\frac{S^2(\sqrt{2n})}{n^3} &= \displaystyle \frac{2}{\pi} \int_0^{\pi} x \ dx \\ \displaystyle \frac{8}{9}\pi + \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{S^2(\sqrt{2n})}{n^3} &= \displaystyle \pi \end{array}$

Therefore, we finally have

$\sum_{n=1}^{\infty} \frac{S^2(\sqrt{2n})}{n^3} = \boxed{\dfrac{\pi^2}{18}}$