I'm working on understanding the convergence properties of certain improper integrals and encountered the following integral:
$$\int_{0}^{\infty} \left| \frac{\cos(x)}{\sqrt{x}} \right| \, dx$$
I believe this integral diverges when considering the absolute value, but I'm struggling to formally prove it. My initial thought was to apply comparison tests or limit comparison tests since the absolute value removes the oscillatory nature of the cosine function, making direct integration challenging. However, I'm unsure how to select an appropriate function to compare it to, especially near the singularity at x=0 and as x approaches infinity.
$$\int_0^{\infty }|\cos x|\frac{dx}{\sqrt{x}}\geq \sum_{k=1}^{\infty}\int_{k\pi}^{k\pi+\frac{\pi}{4}}|\cos x|\frac{dx}{\sqrt{x}}$$$$=\int_0^\frac{\pi}{4}\sum_{k=1}^{\infty}|\cos x|\frac{dx}{\sqrt{x+k\pi}}\geq \frac{\sqrt{2}}{2}\int_0^\frac{\pi}{4}\sum_{k=1}^{\infty}\frac{dx}{\sqrt{x+k\pi}}=\infty$$