Let the function $f_n\in L^1(\mathbb R)\cap C_0(\mathbb R)$ be defined for $n\in\mathbb N$ by $$f_n(x):=\left(\frac{x+i}{x-i}\right)^n\frac{1}{(x-i)^2}\,.$$ Then its Fourier transform is always of the form $$\hat{f_n}(x)=q_n(x)e^{-x}\theta(x)\,,$$ for a polynomial $q_n$ of degree at most $n+1$ and $\theta$ the Heaviside step function. As Jason pointed out in the comments, $q_n(x)=-\sqrt{2\pi}\int_0^x L_n(2t)dt$, with $L_n$ the $n$th Laguerre polynomial.
Clearly, $\|\hat{f_n}\|_1<\infty$. However, is $\|\hat{f_n}\|_1$ bounded as a function in $n$?
Why I need this:
I'm studying regularity properties of functions on $\mathbb R$ obtained via the Cayley transform from functions on the circle. I found out that the problem of integrability of the Fourier transform of a large class of functions can be reduced this very concrete question.
What I've tried:
I tried to use the general estimate $\|\hat{f_n}\|_1\leq c(\|f_n\|_2+\|f_n'\|_2)$, but unfortunately $\|f_n'\|_2$ blows up as $n\to\infty$. I thought about using complex analysis, like e.g. contour integration, but I am not very well versed in that subject. I also used Wolframalpha to determine the polynomials $q_n$, and subsequently determining $\|\hat{f_n}\|_1$ for small $n$. I got (up to a possible factor of $\sqrt{2\pi}$):
$\|\hat{f_0}\|_1=1$
$\|\hat{f_1}\|_1=1.20728$
$\|\hat{f_2}\|_1=1.37548$
$\|\hat{f_3}\|_1=1.52031$
$\|\hat{f_4}\|_1=1.64932$
before my Wolframalpha computation time was exceeded. It looks as though this sequence might stay below 2 (or some larger number) but I couldn't prove this.
This may be coming late to the game at this point, but here's how we can work through the details to show that the $L^1$ norms of the $\hat{f_n}$ are unbounded. I won't get into a justification of the formula for the $q_n$ in terms of the Laguerre polynomials. And I want to note that I'm drawing heavily on Szegö's Orthogonal Polynomials, and will be mirroring the notation there quite a bit.
Step 1: Converting to generalized Laguerre polynomials
To begin, let's start with the formula for the Laguerre polynomails themselves: we have $$ L_n(x) = \sum_{k = 0}^{n} \binom{n}{k} \frac{(-1)^k}{k!} x^k. $$ Using the relationship $q_n(x) = -\sqrt{2 \pi} \int_{0}^{x} L_n(2t) \, dt$, but dropping the constant factor of $-\sqrt{2 \pi}$ for simplicity, we get $$ q_n(x) = \sum_{k = 0}^{n} \binom{n}{k} \frac{(-1)^k}{(k+1)!} 2^k x^{k+1}. $$ With a little extra algebra, we can express the $q_n$ in terms of the generalized Laguerre polynomials $L_n^{(\alpha)}$ as follows: $$ q_n(x) = \frac{x}{n+1} L_n^{(1)}(2x). $$
Step 2: Asymptotics
Now we can directly use the Plancherel-Rotach asymptotics for the $L_n^{(\alpha)}$ as developed in Orthogonal Polynomials. These asymptotics give an approximation for $L_n^{(\alpha)}(x)$ for $x_0 \leq x \leq (4 - \eta)n$, where we can think of $x_0$ and $\eta$ as fixed, small positive numbers. To re-state these asymptotics, we let $l_n = (4n + 4)^{1/2}$ and also introduce auxiliary variables $\xi$ and $\phi$, related to $x$ and to each other by the relationship $x = \xi^2 = l_n^2 \cos^2 \phi$. Then \begin{align*} e^{-x/2} L_n^{(1)}(x) &= e^{-\xi^2 / 2} L_n^{(1)}(\xi^2) \\ &= (-1)^n (\pi \sin \phi)^{-1/2} \xi^{-3/2} n^{1/4} \\ & \qquad \times \left\{ \sin \left( (n + 1) (\sin 2 \phi - 2 \phi) + \frac{3 \pi}{4} \right) + O(\xi^{-1} l_n^{-1}). \right\} \end{align*} For the $q_n$, we therefore obtain \begin{align*} e^{-x/2} q_n(x/2) &= \frac{x}{2n + 2} e^{-x/2} L_n^{(1)}(x) \\ &= 2 l_n^{-2} (-1)^n (\pi \sin \phi)^{-1/2} \xi^{1/2} n^{1/4} \\ & \qquad \times \left\{ \sin \left( (n + 1) (\sin 2 \phi - 2 \phi) + \frac{3 \pi}{4} \right) + O(\xi^{-1} l_n^{-1}) \right\} \\ &= g_n(x/2) + R_n(x/2), \end{align*} where again $x = \xi^2 = l_n^2 \cos^2 \phi$ and the range of $x$-values where this holds is $x_0 \leq x \leq (4 - \eta) n$.
Now we turn to estimating $$ \int_{x_0 / 2}^{(2 - \eta/2) n} | \hat{f_n}(y) | \, dy = \frac{1}{2} \int_{x_0}^{(4 - \eta) n} | e^{-x/2} q_n(x/2) | \, dx $$ from below.
Step 3: The remainder term
We begin with the remainder term $R_n$. Using $C$ to denote a constant that may vary from line to line but which is independent of $n$, we have via the change of variable $t = \cos \phi = \xi / l_n$: \begin{align*} I_R &= \int_{x_0}^{(4 - \eta)n} | R_n(x/2) | \, dx \\ &\leq C l_n^{-3} n^{1/4} \int_{x_0}^{(4 - \eta)n} (\sin \phi)^{-1/2} \xi^{-1/2} \, dx \\ &\leq C l_n^{-3} n^{1/4} \int_0^1 (1 - t^2)^{-1/4} (l_n t)^{-1/2} \cdot 2 l_n^2 t \, dt \\ &= C l_n^{-3/2} n^{1/4} \int_0^1 (1 - t^2)^{-1/4} t^{1/2} \, dt \\ &= C l_n^{-3/2} n^{1/4}. \end{align*} So $I_R = O(n^{-1/2})$.
Step 4a: The main term - changes of variable
The main term with $g_n$ is more involved. As a first step, we carry out a change of variable to $\phi$ in the integral: \begin{align*} I_g &= \int_{x_0}^{(4 - \eta)n} | g_n(x/2) | \, dx \\ &= C l_n^{-2} n^{1/4} \int_{x_0}^{(4 - \eta)n} (\sin \phi)^{-1/2} \xi^{1/2} \left| \sin \left( (n + 1) (\sin 2 \phi - 2 \phi) + \frac{3 \pi}{4} \right) \right| \, dx \\ &= C l_n^{1/2} n^{1/4} \int_{a_n}^{b_n} (\sin \phi)^{1/2} (\cos \phi)^{3/2} \left| \sin \left( (n + 1) (\sin 2 \phi - 2 \phi) + \frac{3 \pi}{4} \right) \right| \, d\phi, \end{align*} where $a_n = \cos^{-1} (\sqrt{(4 - \eta)n}/l_n)$ and $b_n = \cos^{-1} (\sqrt{x_0}/l_n)$. Note that as $n \to \infty$, we have $a_n \to \cos^{-1} (\sqrt{1 - \eta/4})$ and $b_n \to \pi / 2$.
We can additionally change variable via $s = \sin 2 \phi - 2 \phi$; this transformation doesn't permit an entirely closed-form expression for the new integrand, but we can still write $$ I_g = C l_n^{1/2} n^{1/4} \int_{c_n}^{d_n} \tilde{g}(s) \left| \sin \left( (n + 1) s + \frac{3 \pi}{4} \right) \right| \, ds $$ for an appropriate function $\tilde{g}$ and numbers $c_n$ and $d_n$, where $c_n \to -\pi$ and $d_n \to d = \sqrt{\eta (1 - \eta/4)} - 2 \cos^{-1} (\sqrt{1 - \eta/4}) < 0$.
Step 4b: The main term - Féjer's lemma
We can now apply Fejér's lemma to understand the limiting behavior of the remaining integral. This lemma states that if $h_1 \in L^1([0, \pi])$, and if $h_2 \in L^{\infty}(\mathbb{R})$ is $\pi$-periodic, then $$ \lim_{n \to \infty} \int_{0}^{\pi} h_1(s) h_2(ns) \, ds = \left( \frac{1}{\pi} \int_{0}^{\pi} h_2(x) \, ds \right) \left( \int_{0}^{\pi} h_1(s) \, ds \right). $$ A straightforward corollary is that if $h_1^{(n)} \to h_1$ in $L^1([0, \pi])$, then $$ \lim_{n \to \infty} \int_{0}^{\pi} h_1^{(n)}(s) h_2(ns) \, ds = \left( \frac{1}{\pi} \int_{0}^{\pi} h_2(s) \, ds \right) \left( \int_{0}^{\pi} h_1(s) \, ds \right). $$
In our problem, the role of $h_2$ is taken by $| \sin (\cdot + 3\pi / 4) |$, and $$ h_1^{(n)} = \chi_{[c_n, d_n]} \tilde{g} \to \chi_{[-\pi, d]} \tilde{g} = h_1. $$ We can therefore conclude that \begin{align*} \lim_{n \to \infty} &\int_{c_n}^{d_n} \tilde{g}(s) \left| \sin \left( (n + 1) s + \frac{3 \pi}{4} \right) \right| \, ds \\ &= \frac{2}{\pi} \int_{-\pi}^{d} \tilde{g}(s) \, ds \\ &= \frac{2}{\pi} \int_{\cos^{-1} (\sqrt{1 - \eta/4})}^{\pi / 2} (\sin \phi)^{1/2} (\cos \phi)^{3/2} \, d\phi \\ &> 0. \end{align*}
So concluding with the main term, we can say that $I_g \geq C n^{1/2}$ for sufficiently large $n$, where $C > 0$ does not depend on $n$.
The lower bound for $I_g$ and the upper bound for $I_R$ now imply that the norms $\| \hat{f_n} \|_{1}$ are unbounded.