Convergence or divergence of $\int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x$ Is the integral
$$ \int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x $$
convergent or divergent? As the numerator goes to $0$ infinite times I can't apply the comparison test, and I can't seem to find any function to compare the limit. I thought of integrating by parts but I do not know how to prove the integral is divergent. Any hints?

Alternative solution:
Let
$$I_n := \int_3^{\sqrt{2n\pi + 4}} \frac{1 - \cos(4 - x^2)}{x - 2}\mathrm{d} x
\overset{x = \sqrt{y + 4}} = \int_5^{2n\pi} \frac{1 - \cos y}{\sqrt{y + 4} - 2}\, \frac{1}{2\sqrt{y + 4}}\mathrm{d} y.$$
We have
\begin{align*}
 I_n &= \int_5^{2n\pi} \frac{1 - \cos y}{y}\, \frac{\sqrt{y + 4} + 2}{2\sqrt{y + 4}}\mathrm{d} y\\[6pt]
 &\ge \int_5^{2n\pi} \frac{1 - \cos y}{y}\, \frac{\sqrt{y + 4}}{2\sqrt{y + 4}}\mathrm{d} y\\[6pt]
 &= \int_5^{2n\pi} \frac{1 - \cos y}{2y}\mathrm{d} y\\[6pt]
 &\ge \sum_{k=2}^n \int_{(2k-1)\pi - \pi/2}^{(2k-1)\pi + \pi/2} \frac{1 - \cos y}{2y}\mathrm{d} y\\[6pt]
 &\ge \sum_{k=2}^n \int_{(2k-1)\pi - \pi/2}^{(2k-1)\pi + \pi/2} \frac{1}{2y}\mathrm{d} y\\[6pt]
 &= \sum_{k=2}^n \frac12\ln\frac{(2k - 1)\pi + \pi/2}{(2k - 1)\pi - \pi/2}\\
 &\ge \sum_{k=2}^n \frac12 
 \left(1 - \frac{(2k - 1)\pi - \pi/2}{(2k - 1)\pi + \pi/2}\right)\\
 &= \sum_{k=2}^n \frac{1}{4k - 1}\\
    &\ge \sum_{k=2}^n \frac{1}{4k}
\end{align*}
where we have used
$\ln u \ge 1 - u^{-1}$ for all $u\ge 1$ (easy to prove).
Thus, $\int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x$ is divergent.
