Rigorous derivation of the long-time limit of oscillatory integrals I am trying to estimate the following integrals in the limit $t\to+\infty$:
$\displaystyle\int_{-\infty}^{+\infty}\mathrm d\omega\,f(\omega)\frac{1-\cos(\omega t)}{\omega^2}$ and $\displaystyle\int_{-\infty}^{+\infty}\mathrm d\omega\,f(\omega)\frac{\omega t-\sin(\omega t)}{\omega^2}$
Where $f$ is a "nice" (smooth, etc) function. I am quite confident about the following results (numerically it seems to work):
$\displaystyle\int_{-\infty}^{+\infty}\mathrm d\omega\,f(\omega)\frac{1-\cos(\omega t)}{\omega^2}\underset{t\to+\infty}{=}\pi f(0)t+\mathcal P\int_{-\infty}^{+\infty}\mathrm d\omega\,\frac{f'(\omega)}{\omega}+\mathcal O\left(\frac1t\right)$
$\displaystyle\int_{-\infty}^{+\infty}\mathrm d\omega\,f(\omega)\frac{\omega t-\sin(\omega t)}{\omega^2}\underset{t\to+\infty}{=}t\, \mathcal P\int_{-\infty}^{+\infty}\mathrm d\omega\,\frac{f(\omega)}{\omega}-\pi f'(0)t+\mathcal O\left(\frac1t\right)$
Where $\mathcal P$ denotes the Cauchy principal value.
However, I have no clue on how to prove these properties in a somewhat rigorous way; I derived these formulas mostly using my intuition and testing with different explicit expressions for $f(\omega)$.
Does anyone knows whether or not these results have a chance to be right and how to prove it (or maybe give some reference that could be useful)? 
Thanks a lot
 A: 
We assume that $f(\omega)$ is a smooth function with compact support.


Since the kernel, $\frac{1-\cos(\omega t)}{\omega^2}$ is an even function, only the even part, $f_e(\omega)=\frac12(f(\omega)+f(-\omega))$, of $f(\omega)$ contributes to the value of the integral when interpreting it as a Cauchy Principal Value.  
Hence, we can write the integral of interest as
$$\begin{align}
\int_{-\infty}^\infty f(\omega)\frac{1-\cos(\omega t)}{\omega^2}\,d\omega&=2\int_0^\infty f_e(\omega)\frac{1-\cos(\omega t)}{\omega^2}\,d\omega\\\\
&=2f_e(0) \int_0^\infty \frac{1-\cos(\omega t)}{\omega^2}\,d\omega\\\\
&+2\int_0^\infty (f_e(\omega)-f_e(0))\frac{1-\cos(\omega t)}{\omega^2}\,d\omega\\\\
&=\pi f_e(0)t+2\int_0^\infty \frac{f_e(\omega)-f_e(0)}{\omega^2}\,d\omega\tag1\\\\
&+2\int_0^\infty \frac{f_e(\omega)-f_e(0)}{\omega^2}\cos(\omega t)\,d\omega\\\\
&=\pi f_e(0)t+2\int_0^\infty \frac{f_e'(\omega)}{\omega}\,d\omega\\\\
&+2\int_0^\infty \frac{f_e(\omega)-f_e(0)}{\omega^2}\cos(\omega t)\,d\omega\tag2\\\\
&=\pi f_e(0)t+2\int_0^\infty \frac{f_e'(\omega)}{\omega}\,d\omega+O\left(\frac1t\right)\tag3\\\\
&=\pi f(0)t+\text{PV}\int_{-\infty}^\infty\frac{f'(\omega)}{\omega}\,d\omega+O\left(\frac1t\right)\tag4
\end{align}$$

NOTES:
In arriving at $(1)$, we used the fact that 
$$\int_0^\infty \frac{1-\cos(\omega )}{\omega^2}\,d\omega=\frac\pi2$$
In going from $(1)$ to $(2)$ we integrated by parts the first integral on the right-hand side of $(1)$ with $u=f(\omega)-f(0)$ and $v=-\frac1{\omega^2}$.
In writing $(3)$, we appealed to the Riemann-Lebesgue Lemma.
In going from $(3)$ to $(4)$, we used $f_e(0)=f(0)$, and $\text{PV}\int_{-\infty}^\infty \frac{f'(\omega)}{\omega}\,d\omega=2\int_0^\infty \frac{f'_e(\omega)}{\omega}\,d\omega$.

For the second integral of interest, we note that the kernel, $\frac{\omega t-\sin(\omega t)}{\omega^2}$ is an odd function and therefore only the odd part, $f_0(\omega)=\frac12(f(\omega)-f(-\omega))$ contributes to the value of the integral, interpreted as a Cauchy Principal Value.  
Proceeding analogously to the development that led to $(4)$, we have
$$\begin{align}
\int_{-\infty}^\infty f(\omega)\frac{\omega t-\sin(\omega t)}{\omega^2}\,d\omega&=2\int_0^\infty f_o(\omega)\frac{\omega t-\sin(\omega t)}{\omega^2}\,d\omega\\\\
&=2t\int_0^\infty \frac{f_o(\omega)}{\omega}\,d\omega-2f_o'(0)\int_0^\infty \frac{\sin(\omega t)}{\omega}\,d\omega\\\\
&-\int_0^\infty \frac{f_0(\omega)-f'_o(0)\omega}{\omega^2}\sin(\omega t)\,d\omega\\\\
&=2t\int_0^\infty \frac{f_o(\omega)}{\omega}\,d\omega-\pi f_o'(0)+O\left(\frac1t\right)\\\\
&=t\text{PV}\int_{-\infty}^\infty \frac{f(\omega)}{\omega}\,d\omega-\pi f'(0)+O\left(\frac1t\right)\end{align}$$
And we are done!
