Integral of a Fourier transform of a function If we have some function, $f(x)$, are there any tricks to compute an integral of its Fourier transform, $ F(k) $, without first computing the Fourier transform. In other words, if you're given $f(x)$ but don't know $F(k)$, can one compute (numerically if needed):
$$\int_{a}^{b}F(k) dk $$
even when F(k) itself is difficult to find (and probably contains multiple $\delta$-functions)?
 A: $$\int \hat f(\xi) \, d\xi = \left. \int \hat f(\xi) \, e^{-i \xi x} \, dx \right|_{x=0} = \hat{\hat f}(0) = 2\pi \, f(0)$$

The integral with limits can be written
$$\int_a^b \hat f(\xi) \, d\xi = \int \chi_{[a, b]}(\xi) \hat f(\xi) \, d\xi$$
Now, 
$\widehat{\chi_{[a,b]}}(\xi) = (b-a) e^{-i\frac{a+b}{2}\xi} \operatorname{sinc}\left(\frac{b-a}{2}\xi\right),$
$u(x) = \frac{1}{2\pi} \hat{\hat u}(-x),$
and
$\mathscr F\{f(x)\}(-\xi) = \mathscr F\{f(-x)\}(\xi),$
so
$$\begin{align}
\chi_{[a,b]}(\xi) 
& = \frac{1}{2\pi} \mathscr F \left\{ (b-a) e^{i\frac{a+b}{2}x} \operatorname{sinc}\left(-\frac{b-a}{2}x\right) \right\}(-\xi) \\
& = \frac{b-a}{2\pi} \mathscr F \left\{ e^{i\frac{a+b}{2}x} \operatorname{sinc}\left(\frac{b-a}{2}x\right) \right\}(-\xi) \\
& = \frac{b-a}{2\pi} \mathscr F \left\{ e^{-i\frac{a+b}{2}x} \operatorname{sinc}\left(\frac{b-a}{2}x\right) \right\}(\xi) \\
\end{align}$$
This makes
$$
\int_a^b \hat f(\xi) \, d\xi
= \frac{b-a}{2\pi} \int \mathscr F \left\{ e^{-i\frac{a+b}{2}x} \operatorname{sinc}\left(\frac{b-a}{2}x\right) \right\}(\xi) \hat f(\xi) \, d\xi
$$
But $\hat u \hat v = \widehat{u*v}$ so
$$
\int_a^b \hat f(\xi) \, d\xi
= \frac{b-a}{2\pi} \int \mathscr F \left\{ e^{-i\frac{a+b}{2}x} \operatorname{sinc}\left(\frac{b-a}{2}x\right) * f \right\}(\xi) \, d\xi
$$
Now we can use the previous result:
$$
\int_a^b \hat f(\xi) \, d\xi
= (b-a) \left\{ e^{-i\frac{a+b}{2}x} \operatorname{sinc}\left(\frac{b-a}{2}x\right) * f \right\}(0)
$$
We can expand this to an integral using the definition $(u*v)(x) = \int u(y) v(x-y) \, dy$ and $\operatorname{sinc}$ being an even function:
$$
\int_a^b \hat f(\xi) \, d\xi
= (b-a) \int e^{-i\frac{a+b}{2}x} \operatorname{sinc}\left(\frac{b-a}{2}x\right) f(x) \, dx
$$
However, this gives an integral that in general is even worse that directly calculating $\hat f$.
A: If $F(k)$ contains $\delta$'s, you're going to have trouble when some of the $\delta$'s occur at or very close to $a$ or $b$.  But let's ignore that and work formally.  If $\chi$ is the indicator function of the interval $[a,b]$,
$$\int_a^b F(k)\; dk = \langle \chi, F \rangle = \langle \check{\chi}, f \rangle = \int_{-\infty}^\infty \check{\chi}(x)\; f(x)\; dx$$
where the inverse Fourier transform of $\chi$ is
$$ \check{\chi}(x) = \frac{e^{iax} - e^{ibx}}{2\pi x} $$
