Convergence of $\int_{0}^{\pi^{2}}\frac{1}{1-\cos(\sqrt{x})}dx$ I need to compare this integral $\int_{0}^{\pi^{2}}\frac{1}{1-\cos(\sqrt{x})}dx$ to another one in order to ensure its convergence, but I can't find the proper one, because $\frac{1}{1-\sqrt{x}}$ is not continuous.

Because for each small $t $ the series for the cosine is alternating with decreasing absolute values of the term, we have by Leibnitz criterion that $$\tag1 1\cos t\leq1-\frac  {t^2}2. $$ More precisely, using Taylor's remainder we have $$\cos t=1-\frac {t^2}2+\frac { (\cos c)\, t^4}{24}$$ for $c $ between $0$ and $t $. If $0 <t <\frac\pi2$, then the error term is positive and we get $(1) $. Then  $$\frac1 {1-\cos\sqrt x}\geq\frac1 {(\sqrt x)^2/2}=\frac2x, $$and  $$\int_0^{\pi^2}\frac1 {1-\cos\sqrt x}\,dx\geq\int_0^{\pi/4}\frac1 {1-\cos\sqrt x}\,dx\geq\int_0^{\pi/4}\frac 2 {x}\,dx=\infty. $$
