Recently I encouter an integral of this form $$\int_0^\infty \frac{k^2}{1+k^2}\cos(ak)dk$$ with $a$ a parameter we can tune. As $k\to\infty$ the fraction in front approaches $1$ and the integrand becomes simply $\cos(ak)$, whose integral does not exist. Intuitively the integral should oscillate as a function of $t$ if we set the upper bound to be $t$ $$f_a(t)=\int_0^t\frac{k^2}{1+k^2} \cos(ak)dk$$
When I play around with this quantity numerically, I notice that the range of oscillation (the amplitude) seems to depend on $a$. The oscillation seems to have an amplitude of order $O(1/a)$, which I believe is due to the fact that $\cos(ak)$ switches sign whenever $k$ changes by $\pi/a$. However, this is not a rigorous proof. Another interesting thing I notice is that for small $a$, we have
$$\lim_{a\to0}\int_0^{\pi/a}\frac{k^2}{1+k^2}\cos(ak)dk=-\pi/2$$
To prove these properties, I tried extending $k$ into complex numbers and do something like
$$\frac{1}{2}\int_{-\pi/a}^{\pi/a}\frac{k^2}{1+k^2}e^{iak}dk$$ and make use of the residue theorem. However due to non-convergence it seems impossible to simply extend the integration bounds to $\infty$. Does any has any hint on how to handle integrals like this?
Note that \begin{align*} \int_0^t {\frac{{k^2 }}{{1 + k^2 }}\cos (ak){\rm d}k} &= \frac{{\sin (at)}}{a} - \int_0^t {\frac{{\cos (ak)}}{{1 + k^2 }}{\rm d}k} \\& = \frac{{\sin (at)}}{a} - \int_0^{ + \infty } {\frac{{\cos (ak)}}{{1 + k^2 }}{\rm d}k} + \mathcal{O}\!\left( {\frac{1}{t}} \right) \\& = \frac{{\sin (at)}}{a} - \frac{\pi }{2}{\rm e}^{ - \left| a \right|} + \mathcal{O}\!\left( {\frac{1}{t}} \right), \end{align*} for large positive $t$, uniformly for $a\neq 0$. In particular, $$ \int_0^{\pi /a} {\frac{{k^2 }}{{1 + k^2 }}\cos (ak){\rm d}k} = - \frac{\pi }{2}{\rm e}^{ - \left| a \right|} + \mathcal{O}(a) = - \frac{\pi }{2} + \mathcal{O}(a) $$ as $a\to 0$.