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 $\lim_k\Lambda_k x=0$.
1. BOTTOM LINE UP FRONT: Treat it as a distribution.
Since the signum function is not integrable on $\mathbb{R}$, it may be useful to view it as a tempered distribution.
Such "generalized functions" are bounded linear functionals on a class of very-well-behaved functions called Schwartz functions. One of Laurent Schwartz's achievements was finding a collection $\mathcal{S}$ of functions on $\mathbb{R}^n$ such that the set of Fourier transforms of these functions is $\mathcal{S}$ itself. That put the original functions and their Fourier transforms on equal footing.
2. Fourier transform of a distribution
Why is this useful? It means that every tempered distribution has a Fourier transform that is also a tempered distribution. It also provides some useful notation to derive expressions and properties of the Fourier transform of a known tempered distribution.
Given any distribution
$\mathsf{T}$ we write the result of applying it to a Schwartz function
$\varphi$ as
$\left<\mathsf{T},\varphi\right>$, but it is to be understood that this is not an inner product of two objects of the same kind. The Fourier transform of the distribution
$\mathsf{T}$ is the distribution
$\widehat{\mathsf{T}}$ for which
\begin{equation}
\left<\widehat{\mathsf{T}},\varphi\right> = \left<\mathsf{T},\widehat{\varphi}\right>
\end{equation}
for every
$\varphi\in\mathcal{S}$, where
$\widehat{\varphi}$ is the Fourier transform of
$\varphi$. Since
$\varphi\in\mathcal{S}$,
$\widehat{\varphi}\in\mathcal{S}$, too.
3. Fourier transform of signum
How is this related to the signum function? If
$\mathsf{T}$ is the signum function viewed as a distribution, then
\begin{equation}
\left<\mathsf{T},\varphi\right> = \int\textrm{sgn}(x)\varphi(x)dx.
\end{equation}
The Fourier transform of this distribution satisfies (or is defined by)
\begin{equation}
\begin{split}
\left<\widehat{\mathsf{T}},\varphi\right> &=~
\left<\mathsf{T},\widehat{\varphi}\right>\\
&=~ \int\textrm{sgn}(x)\widehat{\varphi}(x)dx\\
&=~ -\int_{-\infty}^{0}\widehat{\varphi}(x)dx + \int_{0}^{\infty}\widehat{\varphi}(x)dx.
\end{split}
\end{equation}
4. Changing order of integration
Let's consider the integral for positive reals. The very good behavior of
$\varphi$ allows changing orders of integration in many, many situations.
\begin{equation}
\begin{split}
\int_{0}^{\infty}\widehat{\varphi}(x)dx
&=~
\int_{0}^{\infty}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~
\lim_{R\to\infty}\int_{0}^{R}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~
\lim_{R\to\infty}\int\left[\int_{0}^{R}e^{-ixk}dx\right]\varphi(k)dk
\end{split}
\end{equation}
We do something very similar for the negative reals.
\begin{equation}
\begin{split}
-\int_{-\infty}^{0}\widehat{\varphi}(x)dx
&=~
-\int_{-\infty}^{0}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~\lim_{R\to\infty}-\int_{-R}^{0}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~
\lim_{R\to\infty}-\int\left[\int_{-R}^{0}e^{-ixk}dx\right]\varphi(k)dk
\end{split}
\end{equation}
We now address the sum of the $R$-dependent integrals.
\begin{equation}
\begin{split}
\int_{0}^{R}e^{-ikx}dx - \int_{-R}^{0}e^{-ikx}dx
&=~
\left.\frac{e^{-ikx}}{-ik}\right|_{x=0}^{x=R}
-
\left.\frac{e^{-ikx}}{-ik}\right|_{x=-R}^{x=0}\\
&=~
\frac{1 - e^{-ikR}}{-ik}
-
\frac{e^{ikR} - 1}{-ik}\\
&=~
\frac{e^{ikR} + e^{-ikR}}{ik}
-
\frac{2}{ik}
\end{split}
\end{equation}
5. Singularity at $k = 0$; Riemann-Lebesgue Lemma
The
$k$ in the denominator will be a problem at
$k=0$. But we know that the original integrals coverge. We must consider the new one as the limit of integrals from
$\epsilon$ to
$\infty$ and from
$-\infty$ to
$-\epsilon$.
\begin{equation}
\int_{|k|>\epsilon}\frac{e^{ikR} + e^{-ikR}}{ik}\varphi(k)dk
=
\int 1_{\{k:|k|>\epsilon\}}(k)\frac{\varphi(k)}{ik}\left(e^{ikR} + e^{-ikR}\right)dk
\end{equation}
For each $\epsilon >0$, the function $1_{\{k:|k|>\epsilon\}}(k)\frac{\varphi(k)}{ik}$ is in $L^1(\mathbb{R})$, so this integral is that function's Fourier transform evaluated at $\omega = R$ plus the same Fourier transform evaluated at $\omega = -R$. The Riemann-Lebesgue Lemma shows that if $f\in L^1(\mathbb{R})$, then $\lim_{|R|\to\infty}\widehat{f}(R) = 0$. Hence, these $R$-dependent terms vanish as $R\to\infty$.
It is worth noting that this shows that we must take the $R$-limit first and then take the $\epsilon$-limit. The opposite order would not work.
6. Cauchy Principal Value
We are left with
\begin{equation}
\lim_{\epsilon\to 0}2i\int_{|k|>\epsilon}\frac{\varphi(k)}{k}dk.
\end{equation}
This is the
Cauchy Principal Value of this integral. This shows that we must interpret the Fourier transform of the signum function very carefully, but we
can do it in the sene of distributions: if
$\textrm{sgn}(x)$ is the signum of
$x$, then
\begin{equation}
\widehat{\textrm{sgn}}(k) = 2i~\mathsf{PV}\left(\frac{1}{k}\right).
\end{equation}
Best Answer
I'll abuse notation and write $u(x) \in L^1(\mathbb R)$ for an expression $u(x)$ instead of writing the more correct $x \mapsto u(x) \in L^1(\mathbb R)$
Suppose that $\hat f(\xi) \in L^1(\mathbb R)$. Then also $\hat f(\xi) e^{i \xi x} \in L^1(\mathbb R)$.
By the Fourier inversion formula, $$f(x) = \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x} d\xi$$
We want to show that for any $x_0 \in \mathbb R$ we have $\lim_{x \to x_0} f(x) = f(x_0)$.
It's enough to show that $\lim_{x_n \to x_0} f(x_n) = f(x_0)$ for any sequence $x_n \to x_0$.
Let $\hat f_n(\xi) = \hat f(\xi) e^{i \xi x_n}$ and $g(\xi) = |\hat f(\xi)|$. Since $e^{i \xi x}$ is continuous in $x$, we have pointwise convergence, i.e. $\hat f_n(\xi) \to \hat f(\xi)$ for every $\xi \in \mathbb R$. Also, for all $n$, $|\hat f_n(\xi)| \leq g(\xi)$ where $g(\xi) = |\hat f(\xi)| \in L^1(\mathbb R)$.
By the dominated convergence theorem then $\int \lim_{n \to \infty} \hat f_n(\xi) d\xi = \lim_{n \to \infty} \int \hat f_n(\xi) d\xi$, i.e. $$f(x) = \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x} d\xi = \frac{1}{2\pi} \lim_{n \to \infty} \int \hat f_n(\xi) d\xi = \lim_{n \to \infty} \frac{1}{2\pi} \int \hat f(\xi) e^{i \xi x_n} d\xi = \lim_{n \to \infty} f(x_n)$$
Thus, $f$ is continuous.