A tricky integral using Fourier Transform and Dirac-functions

Question

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 :)

Answer 1

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

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

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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 $(2)$.

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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 $$\int_0^{t} \int_0^\infty f'(t')\omega^2 \cos(\omega(t'-t))\,d\omega\,dt'=-\frac\pi2 f'''(t)$$

Answer 2

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.