integral and sum10

I will start by focusing on the integral $\int_0^{\infty } \frac{\sin \left(x^{k^2}\right)dx}{x}$ . It is pretty clear that this integral is somehow related to the well known integral $\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \frac{\pi}{2}$ and in order to solve this problem I will find what this relationship must be.

Going back to the integral we are trying to solve, consider the substitution $u = x^{k^2}$ . This substitution will preserve the current limits of integration so we don't have to worry about that. Now, $du = k^2 x^{k^2 - 1} dx = k^2 x^{k^2} \frac{dx}{x} = k^2 u \frac{dx}{x}$ and therefore $\frac{dx}{x} = \frac{du}{k^2 u}$ . So the original integral must be equal to $\int_{0}^{\infty} \frac{\sin(u) du}{k^2 u} = \frac{\pi}{2k^2}$ . Plugging this into the series we now have to compute $\sum_{k=1}^{\infty} \frac{\pi}{2k^2} = \frac{\pi}{2} \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi}{2} \frac{\pi^2}{6} = \frac{\pi^3}{12}$ .