A tricky integral using Fourier Transform and Dirac-functions I need to calculate the following integral:
$$
\boxed{I= \int_{0^+}^{t} \int_0^\infty f'(t')\,  \omega^2 \cos(\omega(t'-t))\, d\omega\, dt'}
$$
where $t>0$, $t' \in (0,t]$ and $f'(x)$ is the derivative of the function $f$.
My attempt:
First I define
$$
A= \int_0^\infty \omega^2\cos(\omega \, a)  \, d\omega
$$
using the fact that $\omega^2\cos(\omega \, a)$ is even in $\omega$,
$$
A= \frac{1}{2} \int_{-\infty}^{\infty} \omega^2\cos(\omega \, a) d\omega
$$
expressing the cosine as a sum of exponentials,
$$
A= \frac{1}{4}\int_{-\infty}^{\infty} \omega^2 \left[ e^{-i \omega (-a)}+e^{-i \omega a}\right] d\omega
$$
The Fourier transform of a polynomial is given in:  https://en.wikipedia.org/wiki/Fourier_transform#Distributions,_one-dimensional
$$
\int_{-\infty}^{\infty} x^n e^{-i \nu x} dx = 2 \pi i^n \delta^{(n)}(x)
$$
where $\delta^{(n)}(x)$ is the n-th derivative of the Dirac-delta distribution.
Therefore,
$$
A= - \frac{\pi}{2} \left[ \delta^{(2)}(-a) + \delta^{(2)}(a)\right].
$$
In this page: https://mathworld.wolfram.com/DeltaFunction.html we can check equation (17), which states that $x^n \delta^{(n)}(x)= (-1)^n n! \delta(x)$, which we can use to deduce that $\delta^{(2)}(x)$ is "even". We conclude that
$$
A= \int_0^\infty \omega^2\cos(\omega \, a)  \, d\omega = - \pi \delta^{(2)}(a).
$$
Using this result in $I$,
$$
I= - \pi \int_{(0^+,t]} f'(t') \delta^{(2)}(t'-t) dt'
$$
after this, I am stuck, I am supposed to prove that
$$
I= -\pi \left[ - f''(t) \delta(0) + \frac{1}{2} f'''(t)\right]
$$
but can not figure how.
Thank you for reading :)
 A: 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 $$\int_0^{t} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'=-\frac\pi2 f'''(t)$$

A: For what it's worth, OP's integral
$$\begin{align}I~=~&\ldots\cr
~=~&-\pi\int_0^t \!\mathrm{d}t^{\prime}~f^{\prime}(t^{\prime})~ \delta^{\prime\prime}(t^{\prime}\!-\!t)\cr
~=~&-\pi{\rm sgn}(t)\int_{\mathbb{R}} \!\mathrm{d}t^{\prime}~1_{[\min(0,t),\max(0,t)]}(t^{\prime})~f^{\prime}(t^{\prime})~ \delta^{\prime\prime}(t^{\prime}\!-\!t)\cr
~\stackrel{t>0}{=}~&-\pi\int_{\mathbb{R}} \!\mathrm{d}t^{\prime}~1_{[0,t]}(t^{\prime})~f^{\prime}(t^{\prime})~ \delta^{\prime\prime}(t^{\prime}\!-\!t)
\end{align}$$
is ill-defined in distribution theory.
