I'm asked to find the explicit form of $(eq.1)$. My first attempt was an heuristic one, given by knowing few basic properties of fourier transform and a remarkable transform (see third column on the rule: 101/106 and 208 in https://en.wikipedia.org/wiki/Fourier_transform#Tables_of_important_Fourier_transforms). Anyway, it gives the following chain of equalities: $(eq.2)$. This result is correct (i have the final solutions of exercises by the book). Furthermore, it suggest me that every functions like the one in $(eq.3)$ can be evaluate using the same method. However i cannot use those remarkable results and i should solve the problem using explicitely the definition, that is solving explicitely something like the integral in $(eq.4)$. Notice that i have changed the function to integrate since i believe that the original function is part of a more general class that can still be analyzed in similar manner. At this point i got stuck, untill i found the residue theorem. In particular, i found this good example from wikipedia that i was trying to adapt to my situation (https://en.wikipedia.org/wiki/Residue_theorem#An_integral_along_the_real_axis). Then i followed the same steps (integration along the two semicircles enclosing each of the poles: $\pm \, ia$, and the calculation of the respective residues). But in my case i can't really understand the following:
- At some point i get the expression in $(eq.5)$. Now, the claim is that the last integral should be null, but i managed to show that the estimation done by wikipedia holds only for $(\beta =0)$.. it is not the case! Is there a way to demonstrate that this integral goes to zero?
- In the case of $(\beta =1)$ i did the desperate act to suppose that the last integral is (indeed) zero.. so i get the result in $(eq.6)$, where the module shows up from evaluating both the cases: $(x\gt0, \; x\lt0)$. From this point ahead i can continue, obtaining the result in $(eq.7)$. We can see that the result itself is good, but there isn't signum function like there was in the $(eq.2)$.. and it should be there.. how could the signum function appear following that method?
$$\mathcal{F^{-1}}\left[-\frac{ik}{\sqrt{2\pi}\text{ }(k^2+4)} \right] \qquad (1)$$
$$\mathcal{F^{-1}}\left[-\frac{ik}{\sqrt{2\pi}\text{ }(k^2+4)} \right] = -\frac{1}{4} \mathcal{F^{-1}}\left[\frac{4(ik)}{\sqrt{2\pi}\text{ }(k^2+4)} \right] \\[6ex] \text{where:} \quad \left[ \frac{4(ik)}{\sqrt{2\pi}\text{ }(k^2+4)} \right] = \mathcal{F} \left[ \frac{d}{dx}\text{ }e^{-2|x|} \right] = (ik)\text{ } \mathcal{F} \left[e^{-2|x|} \right] \\[6ex] \text{and then:} \quad \mathcal{F^{-1}}\left[-\frac{ik}{\sqrt{2\pi}\text{ }(k^2+4)} \right] = -\frac{1}{4}\text{ } \mathcal{F^{-1}} \left(\mathcal{F} \left[ \frac{d}{dx}\text{ }e^{-2|x|} \right]\right) = \\[6ex] = -\frac{1}{4}\text{ }\left[ \frac{1}{4}\text{ }e^{-2|x|} \right] = \frac{1}{2}\text{ }\operatorname{sgn}(x)\text{ }e^{-2|x|}\qquad (2)$$
$$f(k) = \frac{(ik)^\beta}{k^2 + a^2} \quad \text{where} \; \beta \in \mathbb{N}, \; \text{and} \; a \in \mathbb{R^+} \qquad (3)$$
$$\int_{-\infty}^{+\infty} \frac{(ik)^\beta}{k^2 + a^2} \text{ } e^{ikx} \text{ } dk \qquad (4)$$
$$\int_{\operatorname{straight}} \frac{(ik)^\beta \text{ }e^{ikx}}{k^2 + a^2}\text{ }dk = 2\pi i \left[\operatorname{Res}_{z=ia}\right] - \int_{arc} \frac{(ik)^\beta \text{ }e^{ikx}}{k^2 + a^2}\text{ }dk \quad \text{where} \; \left[\operatorname{Res}_{z=ia}\right]=\frac{1}{2}ie^{-ax} \qquad (5)$$
$$\int_{-\infty}^{+\infty} \frac{ik}{k^2 + a^2} \text{ } e^{ikx} \text{ } dk = 2\pi i \left( \frac{1}{2}\text{ }i\text{ }e^{-a|x|} \right)=-\pi\text{ }e^{-a|x|} \qquad (6)$$
$$ \mathcal{F^{-1}}\left[-\frac{ik}{\sqrt{2\pi}\text{ }(k^2+4)} \right] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} -\frac{ik}{\sqrt{2\pi}\text{ }(k^2+4)} e^{ikx}\text{ }dk \\[6ex] = -\frac{1}{2\pi} \int_{-\infty}^{+\infty} \frac{(ik)\text{ }e^{ikx}}{k^2+4}\text{ }dk = \frac{1}{2}\text{ }e^{-2|x|} \qquad (7)$$
Let $f_\beta(k) =\frac{(ik)^\beta}{k^2+a^2}$, where $\beta\in \mathbb{N}$ and $a>0$. For $\beta=0$, $f\in L^1$ and has Fourier Transform given by
$$\begin{align} \mathscr{F}\{f_0\}(x)&=\int_{-\infty}^\infty f_0(k) e^{ikx}\,dk\\\\ &=\int_{-\infty}^\infty \frac{1}{k^2+a^2} \,e^{ikx}\,dk\\\\ &=\pi\frac{e^{-|ax|}}{|a|} \end{align}$$
For $\beta >0$, $f_\beta(k)$ is locally integrable and yields a tempered distribution $\left(f_\beta\right)_{D}=\left(\frac{(ik)^\beta}{k^2+a^2}\right)_{D}$ such that for any $\phi \in \mathbb{S}$ (i.e., $\phi$ is a Schwarz Space function)
$$\begin{align} \langle \left(f_\beta\right)_{D},\phi\rangle&= \int_{-\infty}^\infty \frac{(ik)^\beta}{k^2+\beta^2} \phi(k)\,dk\tag1 \end{align}$$
Therefore, we find that for $\beta>0$
$$\begin{align} \langle \mathscr{F}\{f_\beta\},\phi\rangle&=\langle f_\beta, \mathscr{F}\{\phi\}\rangle \\\\ &=\int_{-\infty}^\infty \frac{(ik)^\beta}{k^2+a^2}\int_{-\infty}^\infty \phi(x)e^{ikx}\,dx\,dk\\\\ &=\int_{-\infty}^\infty \frac{1}{k^2+a^2}\int_{-\infty}^\infty \phi(x)\frac{d^\beta}{dx^\beta}e^{ikx}\,dx\,dk\\\\ &=\int_{-\infty}^\infty \frac{(-1)^\beta}{k^2+a^2}\int_{-\infty}^\infty \phi^{(\beta)}(x)e^{ikx}\,dx\,dk\\\\ &=(-1)^\beta \frac{\pi}{|a|}\int_{-\infty}^\infty \phi^{(\beta)}(x)e^{-|ax|}\,dx\\\\ &=(-1)^\beta \frac{\pi}{|a|} \langle \phi^{(\beta)},\psi_a \rangle \\\\ &=\frac{\pi}{|a|} \langle \phi,\psi_a^{(\beta)} \rangle \\\\ \end{align}$$
where $\psi_a$ is the distribution $\psi_a(x)=\left(e^{-|ax|}\right)_D$ and $\psi^{(\beta)}$ is the $\beta^{th}$ order distributional derivative of $\psi_a$.
Finally, we conclude that the Fourier transform of $f_\beta$ is given by the distribution
$$\mathscr{F}\{f_\beta\}(x)=\frac\pi{|a|}\frac{d^\beta}{dx^\beta}\left(e^{-|ax|}\right)_D$$
For $\beta=1$, we have
$$\mathscr{F}\{f_1\}(x)=-\pi e^{-|ax|}\text{sgn}(x)$$
For $\beta=2$, we have
$$\mathscr{F}\{f_2\}(x)=\pi |a|e^{-|ax|}-2\pi \delta(x)$$