# 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.

• Try integrating by parts twice. Commented Jul 10, 2020 at 20:25
• @JeanMarie I tried to integrate by parts, but I can't get the right result Commented Jul 11, 2020 at 2:49

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

• I also agree that (1) should be the right answer, but the text I am consulting suggests otherwise. Commented Jul 11, 2020 at 2:51
• Just curious .... What is the answer in the text book ? Commented Jul 11, 2020 at 3:00
• @dufong If we assume that the upper limit is replaced with $t+\varepsilon$, for any $\varepsilon>0$, and then let $\varepsilon\to 0$, all works well. I believe this is the interpretation here. This is not unlike the Laplace transform of the Dirac Delta. If the lower limit is not appropriately interpreted, then the problem you mentioned arises. But in practice, we assume that the lower limit is $0^-$ Commented Jul 11, 2020 at 13:49
• @dufong As I mentioned, the Laplace transform of the Dirac Delta is usually interpreted analogously and is assugned the value of $1$. Commented Jul 11, 2020 at 14:17
• $\delta(0)$ is not a defined object and $\frac1\pi \int_0^\infty d\omega$ is not a distribution. It is also an undefined object. If we replace $t$ with $t^+$ in the outer intergal, then the answer is $-\pi f'''(t)$. If we replace the upper limit with $t^-$ then the answer is $0$. So perhaps the author is using a Principal Value approach here. Commented Jul 11, 2020 at 16:09

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.

• If we assume that the upper limit is replaced with $t+\varepsilon$, for any $\varepsilon>0$, and then let $\varepsilon\to 0$, all works well. I believe this is the interpretation here. This is not unlike the Laplace transform of the Dirac Delta. If the lower limit is not appropriately interpreted, then the problem you mentioned arises. But in practice, we assume that the lower limit is $0^-$ Commented Jul 11, 2020 at 13:51
• This book discusses the Laplace transformation in the sense of distribution, see page 69 (10) Commented Jul 11, 2020 at 15:35
• @MarkViola is correct. The integral must be interpreted as including $t$ in its domain, but not zero. That is what I tried to convey in $t' \in (0,t]$. Commented Jul 11, 2020 at 16:01