While generalizing this result, I succeeded in proving that for $\alpha > 0$, $\beta < 1$ and $1 < 2\alpha + \beta < 3$, we have
\begin{align*} &\int_{0}^{\infty} \left[ \left( \int_{x}^{\infty} \frac{\cos t}{t^{\alpha}} \, dt \right)^{2} + \left( \int_{x}^{\infty} \frac{\sin t}{t^{\alpha}} \, dt \right)^{2} \right] \, \frac{dx}{x^{\beta}} \\ & \hspace{10em} = \frac{\pi}{1-\beta} \frac{\Gamma(2-\alpha-\beta)}{\Gamma(\alpha)} \csc \left( \pi \alpha + \frac{\pi (\beta-1)}{2} \right). \end{align*}
My question is
- Is this a known result?
- My ultimate goal is to examine whether the integral $$ I(\alpha, \beta) := \int_{0}^{\infty} \left[ \left( \int_{x}^{\infty} \frac{\cos t}{t^{\alpha}} \, dt \right)^{2} + \left( \int_{x}^{\infty} \frac{\sin t}{t^{\alpha}} \, dt \right)^{2} \right]^{2} \, \frac{dx}{x^{\beta}} $$ has closed from or not for general $\alpha$ and $\beta$. I know that $$ I \left(\tfrac{1}{2}, 0 \right) = 2\pi (\log 4 - 1) \qquad \text{and} \qquad I \left(1, 0 \right) = \frac{2\pi^{3}}{3}, \tag{1} $$ but I know nothing for the other cases. (Here, the former identity in $(1)$ corresponds to the motivating problem linked above.) Is there any other known result concerning this integral?
A further inspection showed that
\begin{align*} \int_{0}^{\infty} \left( ( 1 - 2C(x))^{2} + (1-2S(x))^{2} \right)^{2} \, dx &= \frac{16}{\pi} \left( 1 - \frac{2\sqrt{2}}{\pi} \log \left( 1 + \sqrt{2} \right) \right) \\ &\approx 1.0516193625061961290 \cdots, \end{align*}
where
$$ C(x) = \int_{0}^{x} \cos \left( \tfrac{\pi t^2}{2} \right) \, dt \quad \text{and} \quad S(x) = \int_{0}^{x} \sin \left( \tfrac{\pi t^2}{2} \right) \, dt $$
are Fresnel integrals. Indeed, this corresponds to
$$ I \left( \tfrac{1}{2}, \tfrac{1}{2} \right) = \sqrt{\pi} \left( 4 \sqrt{2} \pi - 16 \log \left( 1 + \sqrt{2} \right) \right). $$
Note that major inverse symbolic calculators do not yield this result.