If for every $t \in \mathbb{R}$, $\left\vert \int_{\mathbb{R}}f(x) e^{-2 \pi ix \cdot t} dx \right\vert < \infty$, is $f(x) \in L^{1}(\mathbb{R})$? I know that if $f \in L^{1}(\mathbb{R})$, 
$$ \left\vert\int_{\mathbb{R}}f(x) e^{-2 \pi ix \cdot t} dx  \right\vert < \infty,~\forall t \in \mathbb{R}$$
However, I am not sure that if the converse problem holds true. That is
if 
$$\left\vert  \int_{\mathbb{R}}f(x) e^{-2 \pi ix \cdot t} dx \right\vert < \infty,~\forall t \in \mathbb{R}$$
is $f \in L^{1}(\mathbb{R})$?
In another word, does there exist a function $f \not \in L^{1}(\mathbb{R})$, for all $t \in \mathbb{R}$, 
$$\left\vert \int_{\mathbb{R}}f(x) e^{-2 \pi ix \cdot t} dx \right\vert< \infty$$
Thanks a lot.
 A: As other users pointed out, you may want to clarify which definition of integral you are working with.
If you are working with Lebesgue integral, then the question is almost a tautology because $f(x)e^{-2\pi i tx}$ is Lebesgue integrable exactly when $f(x)$ is. This is simply because $|f(x)e^{-2\pi i tx}| = |f(x)|$. So the minimal assumption for $f$ which is required to make sense of $\int_{\Bbb{R}} f(x)e^{-2\pi i t x} \, dx$ already answers your question.
That said, you have to work with other notion of integral in order to make your question meaningful.
For instance, you may abuse the notation so that your integral stands for the Fourier transform as isometry $\mathcal{F} : L^2(\Bbb{R}) \to L^2(\Bbb{R})$. Then the answer is negative as pointed out by @Bungo; the image of $L^1(\Bbb{R})\cap L^2(\Bbb{R})$ under $\mathcal{F}$ is strictly contained in $C_0(\Bbb{R})$, the set of continuous functions on $\Bbb{R}$ which vanishes near $\pm \infty$. Clearly there are tons of $L^2$-functions which do not lie in $C_0(\Bbb{R})$.
Or you may extend the Fourier transform on a subspace of $L_{\mathrm{loc}}^{1}(\Bbb{R})$ by adopting a suitable summability method, but even mild forms of such procedure produce counter-examples. For instance, consider the following extension using improper Lebesgue integral:
$$ \int_{-\infty}^{\infty} f(x)e^{-2\pi i tx} \, dx := \lim_{R\to\infty}\int_{-R}^{R} f(x)e^{-2\pi i tx} \, dx. $$
Then we have examples like
$$ \int_{-\infty}^{\infty} \cos(\pi x^2)e^{-2\pi i t x}\, dx = \frac{1}{\sqrt{2}} \left( \cos(\pi t^2) + \sin (\pi t^2) \right). $$
A: No, take f such as: f(x) = 1 if  -1 < x < 1.
f(x) = 1 / abs(x) elsewhere
The Fourier transform exists and the integral of f over R doesn't.
