Integrability of fourier transform Let $f\in L^1(\mathbb{R})$ such that there exist $R,\delta >0$ for which $f$ is bounded in $[-\delta, \delta]$ and 
$\hat{f}(\xi)\geq 0$ for $|\xi|\geq R$. Then $\hat{f}\in L^1(\mathbb{R})$.
Roughly speaking if we were able to use the inversion formula we would have $\int |\hat{f}|=\int \hat{f}=f(0)$ and we would get the
result from boundedness of $f$. I don't know how to make this idea rigorous or if it is the right way.
 A: Since we do not know a priori that $\widehat{f}\in L^{1}(\mathbb{R})$, you cannot simply use the inverse Fourier transform. Instead, you need to use a more general Fourier inversion formula which convolves $f$ with a "good" kernel $\varphi$ so that the function $\widehat{f\ast\varphi}=\widehat{f}\widehat{\varphi}\in L^{1}(\mathbb{R})$. Moreover, you want the kernel $\varphi$ to be "well-localized" about the origin (e.g. compact support) to take advantage of the boundedness of $f$ here.
The trick to using the nonnegativity hypothesis is that $\left|\widehat{f}\right|-\widehat{f}\in L^{1}(\mathbb{R})$, since it is a bounded function of compact support by hypothesis that $\widehat{f}(\xi)\geq 0$ for $\left|\xi\right|\geq R$. The desired conclusion then follows from a more general result.
In what follows below, my definition of Fourier transform, for $f\in L^{1}(\mathbb{R}^{n})$, is
$$\widehat{f}(\xi)=\int_{\mathbb{R}^{n}}f(x)e^{-2\pi ix\cdot\xi}dx,\qquad \xi\in\mathbb{R}^{n}$$

Proposition. Let $f\in L^{1}(\mathbb{R})$. If
  $\left|\widehat{f}\right|-\widehat{f}\in L^{1}(\mathbb{R})$ and $f$ is bounded in a neighborhood of the origin, then $\widehat{f}\in L^{1}(\mathbb{R})$.

The proof presented below can be found in terser form in the book [R. M. Trigub and E. S. Bellinsky, Fourier analysis and approximation of functions. Kluwer Academic Publishers, Dordrecht, 2004.], but in case you don't have access here it is.
Proof. Recall that the Fourier transform of the characteristic function $\chi_{[-h,h]}$ is given by
$$\widehat{\chi}_{[-h,h]}(\xi)=\dfrac{e^{-2\pi ih\xi}-e^{2\pi ih\xi}}{-2\pi i\xi}=\dfrac{\sin(2\pi h\xi)}{\pi\xi} \tag{1}$$
This function is square integrable, and $\widehat{\chi}_{[-h,h]}=\widehat{\chi_{[-h,h]}\ast\chi_{[-h,h]}}$ by the convolution theorem. Set
$$\varphi_{h}(x):=\dfrac{1}{(2h)^{2}}\chi_{[-h,h]}\ast\chi_{[-h,h]}(x)\tag{2}$$
Using the inequality $\left|\sin\xi\right|\leq\left|\xi\right|$, we have that
\begin{align*}
\int_{\mathbb{R}}\left|\widehat{f}(\xi)\right|\left(\dfrac{\sin 2\pi h\xi}{2\pi h\xi}\right)^{2}d\xi&=\left|\int_{\mathbb{R}}\left[\left|\widehat{f}(\xi)\right|-\widehat{f}(\xi)\right]\left(\dfrac{\sin 2\pi h\xi}{2\pi h\xi}\right)^{2}d\xi+\int_{\mathbb{R}}\widehat{f}(\xi)\left(\dfrac{\sin 2\pi h\xi}{2\pi h\xi}\right)^{2}d\xi\right|\\&\\
&\leq\int_{\mathbb{R}}\left|\left|\widehat{f}(\xi)\right|-\widehat{f}(\xi)\right|d\xi+\left|\int_{\mathbb{R}}\widehat{f}(\xi)\left(\dfrac{\sin 2\pi h\xi}{2\pi h\xi}\right)^{2}d\xi\right|\tag{3}
\end{align*}
Since the integrand of the second term in (3) belongs to $L^{1}$, we can apply Fourier inversion to obtain
\begin{align*}
\int_{\mathbb{R}}\widehat{f}(\xi)\left(\dfrac{\sin 2\pi h\xi}{2\pi h\xi}\right)^{2}e^{2\pi i \xi x}d\xi&=(f\ast\varphi_{h})(x)=\dfrac{1}{(2h)^{2}}\int_{-h}^{h}\int_{-h}^{h}f(x-t-s)dtds,\tag{4}
\end{align*}
where we use the fact that equality holds everywhere since both sides are continuous. For $\left|x\right|\leq\delta/8$ and $0<h\leq\delta/8$ the integrand on the RHS of (4) is bounded by some constant $M>0$, whence
\begin{align*}
\left|\int_{\mathbb{R}}\widehat{f}(\xi)\left(\dfrac{\sin 2\pi h\xi}{2\pi h\xi}\right)^{2}d\xi\right|\leq M\tag{5}
\end{align*}
We conclude from Fatou's lemma that
\begin{align*}
\int_{\mathbb{R}}\left|\widehat{f}(\xi)\right|d\xi=\int_{\mathbb{R}}\liminf_{h\rightarrow 0}\left|\widehat{f}(\xi)\right|\left(\dfrac{\sin 2\pi h\xi}{2\pi h\xi}\right)^{2}d\xi\leq \int_{\mathbb{R}}\left|\left|\widehat{f}(\xi)\right|-\widehat{f}(\xi)\right|d\xi+M\tag{6}
\end{align*}
$\Box$
