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.
 A: Using small angle approximation, $\sin(x)\approx x$.Hence $\cos(x)\approx 1- \frac{x^2}{2}$. So, $\cos(\sqrt{x})\approx 1- \frac{x}{2}$. Decomposing the given integral,
$$\int_{0}^{\pi^2} \frac{1}{1-\cos(\sqrt{x})}dx=\int_{0}^{\delta} \frac{1}{1-\cos(\sqrt{x})}dx+\int_{\delta}^{\pi^2} \frac{1}{1-\cos(\sqrt{x})}dx,$$ where $\delta>0$ is arbitrary small. Now,
$$\int_{0}^{\delta} \frac{1}{1-\cos(\sqrt{x})}dx \approx \int_{0}^{\delta} \frac{1}{1-(1-\frac{x}{2})}dx=2(\ln|x|)_{0}^{\delta},$$ which approaches $\infty$ as $x \rightarrow 0^+$. Hence,the first integral diverges. So does the original one.
A: @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
$$\frac{9595999}{2400000}+\log (160000)\approx 15.98126201088$$
A: 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  $$\int_0^{\pi^2}\frac1 {1-\cos\sqrt x}\,dx\geq\int_0^{\pi/4}\frac1 {1-\cos\sqrt x}\,dx\geq\int_0^{\pi/4}\frac 2 {x}\,dx=\infty. $$
