Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there's no longer a singularity then, we can split the integral in two:
$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega
&=&
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\
&=&
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega -
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;.
\end{eqnarray}$$
Now we can apply different substitutions to the two parts, $\omega'=\omega(x+\frac{1}{2})$ in the first part and $\omega'=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega'}/(i\omega')$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there's a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get
$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega' +
\int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'\;,$$
which is (again using the sufficient decay at infinity)
$$\oint \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'$$
on a contour that encloses the pole at the origin, and hence the value is $2\pi$.
This seems to be true, which I find very surprising. (Edit: I found it surprising because I was forgetting a classical result - see the note a few paragraphs down.)
First note that $f$ being even is irrelevant to the principal value $\lim_{c\to\infty}\int_{-c}^c\hat f$:
If $\lim_{c\to\infty}\int_{-c}^c\hat f$ exists for every even function $f\in L^1$ with $f'\in L^1$ then the same limit exists for every $f\in L^1$ with $f'\in L^1$.
Proof: Given $f,f'\in L^1$, let $$g(t)=\frac12(f(t)+f(-t)).$$
Then $\hat g$ is even and $$\hat g(\xi)=\frac12(\hat f(\xi)+\hat f(-\xi)),$$so $$\int_{-c}^c\hat f=\int_{-c}^c\hat g,$$qed.
It doesn't follow that evenness is irrelevant to the problem; if $f$ is even then $$\lim_{x\to\infty}\lim_{b\to\infty}\int_{-a}^b\hat f=\lim_{c\to\infty}\int_{-c}^c\hat f.$$
Here's the part that surprises me - I would have thought it would sound familiar if true:
Note: No, it's not surprising at all. The corresponding fact for Fourier series follows in half a line from the fact that if $f$ is periodic and has bounded variation then the Fourier series converges to $f$ at every point of continuity. I may as well leave the rest of this here:
Thm. If $f\in L^1$ is absolutely continuous and $f'\in L^1$ then $\lim_{c\to\infty}\int_{-c}^c\hat f=f(0).$
Note of course that's not quite right, there should be a $\sqrt{2\pi}$ somewhere. Anyway,
Let $X$ be the Banach space of all absolutely continuous integrable $f$ with $f'\in L^1$, with norm $$||f||_X=||f||_1+||f'||_1.$$For $c>0$ let $$\Lambda_cf=\int_{-c}^c\hat f.$$
There exists a bounded function $S$ with $$S'(t)=\frac{\sin(t)}t.$$If $f\in X$ then Fubini's theorem plus an integration by parts show that, again omitting irrelevant constants, $$\Lambda_cf=\int f(t)\frac{\sin(ct)}t=\int f(t/c)\frac{\sin(t)}t=-\int\frac{f'(t/c)}c S(t).$$
So $$|\Lambda_cf|\le||S||_\infty\int\frac{|f'(t/c)|}{|c|}=||S||_\infty||f'||_1\le||S||_\infty||f||_X.$$In particular, $$||\Lambda_c||_{X^*}\le||S||_\infty.$$Now if $f\in X$ then $f(0)=\int_{-\infty}^0f'$, so $f\mapsto f(0)$ is a bounded linear functional on $X$. Since $\Lambda_cf\to f(0)$ for all $f$ in a dense subspace and $||\Lambda_c||_{X^*}$ is bounded it follows that $\Lambda_cf
\to f(0)$ for all $f\in X$.
Of course there's nothing special about $0$:
Cor. If $f\in X$ then $\lim_{c\to\infty}\int_{-c}^c\hat f(\xi)e^{it\xi}\,d\xi=f(t)$.
Hint: $\hat f(\xi)e^{it\xi}=\hat g(\xi)$ if $g=???$
Addendum For the benefit of anyone unhappy about the functional analysis above:
Standard Exercise. Suppose $E$ is a Banach space, $\Lambda_n\in E^*$, and $||\Lambda_n||$ is bounded. If $\Lambda_nx\to0$ for all $x$ in some dense subspace of $E$ then $\Lambda_nx\to 0$ for all $x\in E$.
(Apply this with $\tilde\Lambda_cf=\Lambda_cf-f(0)$ above; $\Lambda_cf\to f(0)$ is the same as $\tilde\Lambda_cf\to0$.)
This is nothing but epsilons and deltas. It's also the same as the proof that a uniform limit of continuous functions is continuous (and in fact it follows from that result if you look at it right). Anyway:
Solution Say $||\Lambda_n||\le c$ for all $n$, and say $S$ is the dense subspace in question.
Suppose $x\in E$. Let $\epsilon>0$. Choose $y\in S$ with $$||x-y||<\frac\epsilon{2c}.$$Since $y\in S$ there exists $N$ so $$|\Lambda_ny|<\frac\epsilon2\quad(n>N).$$Now if $n>N$ we have $$|\Lambda_nx|\le|\Lambda_n y|+|\Lambda_n(x-y)|<\frac\epsilon2+||\Lambda_n||\,||x-y||\le\frac\epsilon2+c\frac\epsilon{2c}=\epsilon.$$
The exercise really is a simple consequence of the fact that a uniform limit of continuous functions is continuous (although the argument may be a bit "abstract").
Interesting Solution: Let $K$ be the one-point compactification of $\Bbb N$: $$K=\Bbb N\cup\{\infty\}.$$
Suppose $||\Lambda_n||\le c$ for all $n$ and let $S$ be the given dense subspace of $E$ on which $\Lambda_n\to0$.
Suppose $x\in E$. Choose $y_n\in S$ with $||x-y_n||\to0$. Define $f_n:K\to\Bbb C$ by $$f_n(k)=\begin{cases}\Lambda_ky_n,\quad(k\in\Bbb N),\\0,&(k=\infty).\end{cases}$$Now the fact that $\lim_k\Lambda_ky_n=0$ says precisely that $f_n$ is continuous. If $f$ is defined as $f_n$ was except with $x$ in place of $y_n$ then $$|f_n(k)-f(k)|\le c||y_n-x||\quad(k\in K).$$So $f_n\to f$ uniformly on $K$, hence $f$ is continuous, which says $\lim_k\Lambda_k x=0$.
Best Answer
This is a really nice question, I was surprised after thinking for a couple of hours and none of the standard examples in a harmonic analysis course fitting the description. In any case, I will try to present what can and what cannot be done to the integrability of the Fourier transform of a Lipschitz function.
Notice also that I will be employing standard analysis notation (such as $\lesssim$ and $O(g)$) as well as some standard theorems in Fourier analysis, such as van der Corput's lemma for bounding oscillatory integrals (also known as 'stationary phase method'). If you are not familiar with those, I would gladly refer you to Stein's "Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals", whose chapter VIII contains all that we shall use here.
1.1. If instead of $f$ being Lipschitz, one requires the different condition that $\|f(x+h)-f(x)\|_{L^2(dx)} = O(h^{\alpha}),$ for some $\alpha > 1/2,$ then the same conclusion of Bernstein's theorem holds. That is, if the above condition is satisfied and $f \in L^1(\mathbb{R}),$ then $\widehat{f} \in L^1(\mathbb{R}).$ A proof of this fact can be found, for instance, in the book "Introduction to the Theory of Fourier Integrals", by Titchmarsh.
As a by-product, notice that if $f \in L^1(\mathbb{R})$ is Lipschitz and has compact support, then the condition in Titchmarsh's theorem above is trivially satisfied from the Lipschitz condition, and thus $\widehat{f} \in L^1(\mathbb{R}).$ This shows that, if a counterexample exists, it cannot have compact support.
1.2. On the other hand, notice that $f \in L^1, f' \in L^{\infty}$ implies directly that $f \in L^{\infty},$ and thus, by interpolation, $f \in L^p, \forall p \ge 1.$ By the Hausdorff--Young inequality, we have that $\widehat{f} \in L^q, \forall q \ge 2.$ In fact, this can be significantly improved:
Claim: Let $f \in L^1(\mathbb{R})$ be Lipschitz, i.e., $f' \in L^{\infty}.$ Then $\widehat{f} \in L^p(\mathbb{R}), \forall p > 1.$
Proof: In Grafakos's "Modern Fourier Analysis", the following result can be found as a by-product of a characterization of Lipschitz spaces: 'Let $\Delta_j(f) = (\psi(\xi/2^j) \widehat{f})^{\vee},$ where $\psi$ is a smooth bump supported on $[1/2,4]$ and equal to 1 on $[1,2].$ If $f$ is Lipschitz, then
$$\sup_{j \ge 0} 2^j \|\Delta_j(f)\|_{\infty} < + \infty.'$$
We will not prove this here, but by assuming such a result, we have the following: for $\psi$ as in the statement of the Littlewood-Paley characterization of Lipschitz functions, we have
$$\int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \le \int_{\mathbb{R}} |\psi(\xi/2^j) \widehat{f}(\xi)|^p \psi(\xi/2^j) \, d \xi \lesssim 2^{(1-\frac{p}{2})j} \|\Delta_j(f)\|_2^{p},$$
where in the last inequality we used Hölder and Plancherel. Now, by interpolation we have
$$\|\Delta_j(f)\|_2 \le \|\Delta_j(f)\|_{\infty}^{1/2} \|\Delta_j(f)\|_1^{1/2}.$$
From the definition of $\Delta_j,$ we have $\|\Delta_j(f)\|_1 \lesssim \|f\|_1,$ and from the characterization of Lipschitz functions mentioned before, $\|\Delta_j(f)\|_{\infty} \lesssim 2^{-j}.$ Putting all those considerations together, we obtain
$$ \int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \lesssim 2^{(1-p)j},$$
and the right-hand side above is summable in $j \ge 1$ if $p>1,$ as desired. $\square.$
Thus, the Fourier transform of any such counterexample has to integrate to any order $p>1.$
$$ \widehat{f}(x) = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x).$$
Then we need to prove that
$$f(\xi) = \int_{\mathbb{R}} \frac{e^{i\frac{x^2}{2\log x} - i\xi x}}{x \log x} \psi_0(x) \, dx$$
is Lipschitz and integrable.
2.1. $f$ is Lipschitz: it is not too hard (even though it is a bit technical) to show that $f$ is Lipschitz if and only if the integrals
$$ J_N(\xi) = \int_{10}^N \frac{e^{i\frac{x^2}{2 \log x}}}{\log x} \psi_0(x) e^{-i \xi x} \, dx $$
are uniformly bounded on $\xi, N.$ Define then the phase function $\phi(x,\xi) = \frac{x^2}{2\log x} - \xi x.$ Suppose that, for some $x_{\xi} \in \mathbb{R},$ we have $\partial_x \phi(x_{\xi},\xi) = 0.$ As $\partial_x \phi(x,\xi) = \frac{x}{\log x} + \frac{x}{2 \log^2 x} - \xi,$ then it is also not hard to show that $x_{\xi} \gtrsim |\xi| \log |\xi|.$ We split then the integral defining $J_N(\xi)$ into the one over the interval $[5,N] \setminus [\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^1(\xi),$ and the one over $[\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^2(\xi).$ Thus, $J_N(\xi) = J_N^1(\xi) + J_N^2(\xi).$
For $J_N^1,$ we use that the derivative of the phase satisfies $|\partial_x\phi(x,\xi)| \gtrsim \max\{|\xi|,\frac{x}{\log x}\},$ while for $x$ sufficiently large (10 should do for this) $\partial_{xx}^2 \phi(x,\xi) > 0,$ so we are in position to apply van der Corput's lemma, which then implies that
$$|J_N^1(\xi)| \lesssim \frac{1}{|\xi|}.$$
For the analysis of $J_N^2,$ we notice that the interval $[\xi^{1/2},\xi^{3/2}]$ contains the zero $x_{\xi}$ of $\partial_x \phi(x,\xi),$ and in that interval it also holds that $\partial_{xx}^2 \phi(x,\xi) \gtrsim \frac{1}{\log \xi}.$ Using van der Corput's lemma again, we have that
$$|J_N^2(\xi)| \lesssim \frac{(\log \xi)^{1/2}}{\log \xi} \lesssim \frac{1}{(\log \xi)^{1/2}}.$$
This finishes the case $\xi \gg 1.$ For the small $\xi$ case, the analysis here carries out even easier than what is written here. Thus, $f$ is Lipschitz.
2.2. $f$ is integrable: The idea is basically the same as here, but integrating by parts first. Indeed, we need to prove some uniform estimates on the functions
$$f_N(\xi) := \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0(x) \, dx.$$
Integrating by parts with factors $u = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x)$ and $dv = e^{-i \xi x} dx.$ This gives that $f_N(\xi)$ is equivalently equal to
$$\frac{1}{i\xi}\frac{ e^{i( \frac{N^2}{2\log N} - \xi N)}}{N \log N} + $$ $$ \frac{1}{i \xi} \int_{10}^N \partial_x \left(\frac{x^2}{2\log x}\right) x \log x \frac{e^{i(\frac{x^2}{2\log x} - \xi x)} }{x^2 \log^2 x} \, \psi_0(x) dx + $$ $$ \frac{1}{ i\xi} \int_{10}^N \frac{\log x + 1}{x^2 \log^2 x} e^{i(\frac{x^2}{2 \log x} - \xi x)} \psi_0(x) \, dx + $$ $$ \frac{1}{i\xi} \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0'(x) \, dx =: \frac{1}{i\xi} (I_1(N,\xi) + I_2(N,\xi) + I_3(N,\xi) + I_4(N,\xi)).$$
One notices quite immediately that $I_1(N,\xi) \to 0$ as $N \to \infty,$ so we do not need to worry about that term when taking limits. Also, $\frac{\psi_0'(x)}{x \log x}$ has support on $[10,20]$ and is smooth there, while the derivative of the phase $\partial_x\phi(x,\xi)$ does not vanish for $\xi$ small, and neither does it on the support for $\xi$ sufficiently large. Thus, this term is bounded and decays as fast as one wishes - i.e., $|I_4(N,\xi)| = O_R(\xi^{-R})$ when $\xi \to \infty,$ for any $R > 0.$ Thus $I_4(N,\cdot) \in L^1(\mathbb{R})$ uniformly on $N \to \infty.$
We then analyze the remaining terms $I_2,I_3.$ In fact, the analysis of $I_2$ is much more delicate than that of $I_2,$ so we perform only the former and defer the latter to whoever wishes to fill in the gaps.
Rewrite $I_2(N,\xi)$ as
$$\int_{10}^N \left( \frac{x}{\log x} + \frac{x}{2 \log^2x} \right) \frac{x \log x}{x^2 \log^2 x} e^{i(\frac{x^2}{2\log x} - \xi x)} \psi_0(x) \, dx = $$ $$ = 2 \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{ (\log x)^2} \psi_0(x) dx + \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{(\log x)^3} \psi_0(x) \, dx.$$
These resemble a lot the integrals we had to estimate for proving $f$ is Lipschitz. In fact, the very same argument is applicable in these cases as well, where one obtains (after redoing the same computations with van der Corput and what not)
$$|I_2(N,\xi)| \lesssim \frac{1}{(\log |\xi|)^{3/2}}.$$
As mentioned before, the analysis of $I_3$ is in fact easier, and doing the computations yields
$$|I_3(N,\xi)| \lesssim \frac{1}{|\xi|}.$$
Notice that all such estimates were made uniformly on $N \to \infty.$ Putting all together and taking limits, one obtains that, for $|\xi| \gg 1,$
$$|f(\xi)| \lesssim \frac{1}{|\xi| (\log|\xi|)^{3/2}}.$$
As the right-hand side is integrable and $f$ was already seen to be Lipschitz, this implies $f \in L^1(\mathbb{R}),$ as desired.