Improper Fourier transform The most common way to verify if the Fourier transform of a function $f$ is integrable $(\hat f\in L^1(\mathbb{R}))$ is by proving that the function $f$ is integrable and $f'$, $f''$ are also differentiable. Now, I came across a weaker condition for the Fourier tranform to be improper integrable. The statement was as follows:
If $f:\mathbb{R} \to \mathbb{C}$ is a differentiable, integrable function and even, i.e. $f(x)=f(-x)$, then the Fourier transform $\hat{f}: \mathbb{R} \to \mathbb{C}$ will be an improper integral, i.e. $\hat{f}\chi_{[c,d]}$ is integrable for all $c<d$ and the limit $$\lim_{c \to -\infty}\lim_{d\to +\infty} \int_c^d \hat{f}$$
will exist. Furthemore, would there be a formula for this improper integral in terms of $f$?
Can anyone provide me with a hint for this problem? I have no idea how to start?
 A: 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 $\lim_k\Lambda_k x=0$.
