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.

@SL_MathGuy already showed that the definite integral diverges.
Using the tangent half-angle substitution and one integration by parts, the antiderivative is given by
$$J(x)=\int\frac{dx}{1-\cos \left(\sqrt{x}\right)}=4 \log \left(\sin \left(\frac{\sqrt{x}}{2}\right)\right)-2 \sqrt{x} \cot \left(\frac{\sqrt{x}}{2}\right)$$ and, since $J(\pi^2)=0$
$$\int_\epsilon^{\pi ^2}\frac{dx}{1-\cos \left(\sqrt{x}\right)}=2 \sqrt{\epsilon } \cot \left(\frac{\sqrt{\epsilon }}{2}\right)-4 \log \left(\sin
   \left(\frac{\sqrt{\epsilon }}{2}\right)\right)$$ which, expanded a a series gives
$$-2 \log (\epsilon )+4(1+ \log (2))-\frac{\epsilon }{6}-\frac{\epsilon
   ^2}{240}+O\left(\epsilon ^{3}\right)$$
Computed for $\epsilon=\frac 1 {100}$, the exact result would be
$$\frac{1}{5} \cot \left(\frac{1}{20}\right)-4 \log \left(\sin
   \left(\frac{1}{20}\right)\right)\approx 15.98126201077$$ while the above truncated expansion would give
$$\frac{9595999}{2400000}+\log (160000)\approx 15.98126201088$$
