A tricky integral using Fourier Transform and Dirac-functions

Note that we have

$$\omega^2 \cos(\omega(t'-t))=-\frac{d^2\cos(\omega(t'-t))}{dt'^2}=-\frac{d^2\cos(\omega(t'-t))}{dt^2}$$

Under the assumption that $f(t)$ is a suitable test function, we have in distribution for $t>0$

$$\begin{align} F^+(t)&=\lim_{\varepsilon\to0^+}\int_0^{t+\varepsilon} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'\\\\ &=-\frac12\lim_{\varepsilon\to0^+}\int_0^{t+\varepsilon} f'(t')\frac{d^2}{dt'^2}\int_{-\infty}^\infty \cos(\omega(t'-t))\,d\omega\,dt'\\\\ &=-\frac12 \lim_{\varepsilon\to0^+}\int_0^{t+\varepsilon} f'(t')\frac{d^2}{dt'^2}\left(2\pi \delta(t'-t)\right)\,dt'\\\\ &=-\pi f'''(t)\tag1 \end{align}$$

while

$$\begin{align} F^-(t)&=\lim_{\varepsilon\to0^+}\int_0^{t-\varepsilon} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'\\\\ &=-\frac12\lim_{\varepsilon\to0^+}\int_0^{t-\varepsilon} f'(t')\frac{d^2}{dt'^2}\int_{-\infty}^\infty \cos(\omega(t'-t))\,d\omega\,dt'\\\\ &=-\frac12 \lim_{\varepsilon\to0^+}\int_0^{t-\varepsilon} f'(t')\frac{d^2}{dt'^2}\left(2\pi \delta(t'-t)\right)\,dt'\\\\ &=0\tag2 \end{align}$$

---

Alternatively, we can write

$$\begin{align} \int_0^{t^+} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'&=\frac12\int_0^{t^+} f'(t') \int_{-\infty}^\infty \omega^2 e^{i\omega(t'-t)}\,d\omega\,dt'\\\\ &=\frac12\int_0^{t^+} f'(t') (-2\pi \delta''(t'-t))\\\\ &=-\pi f'''(t) \end{align}$$

in agreement with $(1)$, while

$$\begin{align} \int_0^{t^-} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'&=\frac12\int_0^{t^-} f'(t') \int_{-\infty}^\infty \omega^2 e^{i\omega(t'-t)}\,d\omega\,dt'\\\\ &=\frac12\int_0^{t^-} f'(t') (-2\pi \delta''(t'-t))\\\\ &=0 \end{align}$$

in agreement with $(2)$.

---

---

NOTE:

The notation $$F(t)=\int_0^{t} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'$$is not defined as a distribution since $\delta(x)H(x)$ is not a defined distribution.

However, if we interpret $$F(t)=\int_0^{t} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'$$to be the simple arithmetic average of $F^+(t)$ and $F^-(t)$, then we can write $$\int_0^{t} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'=-\frac\pi2 f'''(t)$$