Inverse Fourier transform of $\frac{\cos(x)}{x^2} $

Question

I am stuck in the integral of an inverse Fourier transform in an exercise of a graduate signals class.

I am trying to compute the inverse Fourier Transform for the signal given below. I tried regular integration techniques to calculate the inverse transform:

$$ x(t) = \frac{1}{2π} \int_{- \infty}^{+ \infty} { Χ(ω) e^{iωt} dω} = \frac{1}{2π} \int_{- \infty}^{+ \infty} { \frac{\cos(ω)}{ω^2} e^{iωt} dω} $$

but they do not work. So what is the right answer?

Summarizing the question:

Find the signal whose Fourier transform is

$$ X(Ω) = \frac{\cos(Ω)}{Ω^2} $$

What is the inverse Fourier transform of the above X(Ω)?

Answer 1

Note that $f(x)=\frac{\cos(x)}{x^2}$ is not locally integrable and is not, therefore, a tempered distribution. We can define, however, a distribution that permits our defining the Fourier Transform. We define the distribution $\left(\psi\right)_{D}=\left(\frac{\cos(x)}{x^2}\right)_{D}$ such that for any $\phi\in \mathbb{S}$,

$$\left(\left(\psi\right)_{D},\phi\right)=\text{PV}\int_{-\infty}^\infty \frac{\cos(x)}{x^2}\left(\phi(x)-\phi(0)-\phi'(0)x\right)\,dx\tag1$$

Equipped with $(1)$, we proceed to find the Fourier Transform of $\left(\psi\right)_{D}$.

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We see that

$$\begin{align} \langle \mathscr{F}\{\left(\psi\right)_{D}\},\phi\rangle&=\langle \left(\psi\right)_{D},\mathscr{F}\{\phi\}\rangle\\\\ &=\text{PV}\int_{-\infty}^\infty \frac{\cos(x)}{x^2}\int_{-\infty}^\infty \phi(k)\left(e^{ikx}-1-ix\right)\,dk\,dx\\\\ &=2\int_0^\infty \frac{\cos(x)}{x^2} \int_{-\infty}^\infty \phi(k)\left(\cos(kx)-1\right)\,dk\,dx\\\\ &=-2\int_{-\infty}^\infty \phi(k)\int_0^\infty \frac{\cos(x)(1-\cos(kx))}{x^2}\,dx\,dk\\\\ &=-\int_{-\infty}^\infty \phi(k)\int_0^\infty \frac{\cos((k+1)x)+\cos((k-1)x)}{x}\,dx\,dk\\\\ &=-\int_{-\infty}^\infty \phi(k)\int_0^\infty \frac{(k+1)\sin((k+1)x)+(k-1)\sin((k-1)x)}{x}\,dx\,dk\\\\ &=-\int_{-\infty}^\infty \phi(k) \frac\pi 2\left((k+1)\text{sgn}(k+1)+(k-1)\text{sgn}(k-1)\right)\,dk \end{align}$$

Therefore, in distribution we see that

$$\mathscr{F}\{\left(\psi\right)_{D}\}=-\frac\pi2 \left((k+1)\text{sgn}(k+1)+(k-1)\text{sgn}(k-1)\right)$$

Answer 2

Here's a method that gives the solution up to an additive constant ($B$ in the result below).

I will skip constant factors. Some of those depend of the used definition of the Fourier transform.

The inverse Fourier transform of $\Omega^2 X(\Omega) = \cos\Omega$ is $$-x''(t) = \delta(t-1)+\delta(t+1).$$ Therefore, $$x(t) = -\frac12 (|t-1|+|t+1|) + At + B,$$ where $A$ and $B$ are constants. However, since $X(\Omega)$ is even, so must be $x(t).$ Therefore $A=0.$