Integral of a Fourier transform of a function

Question

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)?

Answer 1

$$\int \hat f(\xi) \, d\xi = \left. \int \hat f(\xi) \, e^{-i \xi x} \, dx \right|_{x=0} = \hat{\hat f}(0) = 2\pi \, f(0)$$

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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 $\hat f$.

Answer 2

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

$$ \check{\chi}(x) = \frac{e^{iax} - e^{ibx}}{2\pi x} $$