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The Fourier transform and its inverse are definite integrals:

$$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2 \pi i x \xi} dx$$ $$f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi) e^{2 \pi i \xi x} d\xi$$

However I can find no reference to the indefinite counterpart of these integrals, i.e.:

$$\int f(x) e^{-2 \pi i x \xi} dx \text{ }\text{ and } \int \hat{f}(\xi) e^{2 \pi i \xi x} d\xi$$

What is the meaning of each of these indefinite integrals? Can we make statements about $f(x)$ from either of these, and vice versa? Are there any good references discussing these antiderivative functions and their properties?

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Do you know that $$g(x) = \int_{-A}^{A} \hat{f}(\xi) e^{2i \pi \xi x} d \xi = f \ast h(x)$$ where $h(x) = \frac{\sin(\pi x A)}{\pi x}$ is an ideal low-pass filter and $\ast$ is the convolution ?

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    Thank you! This is an interesting fact and one I was not aware of, but I'm not sure how it answers my question about indefinite integrals. – 2017-02-25
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    @KFox It does. What do you get similarly for $\int_A^B \hat{f}(\xi) e^{2i \pi \xi x}d\xi$ ? – 2017-02-25
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    My apologies, but after searching around for several minutes, I can't find the result of $\int_A^B \hat{f}(\xi) e^{2\pi i \xi x} d \xi$ or anything more than a vague connection to the indefinite integral. Would you mind explaining the connection to me? – 2017-02-25
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    @KFox $\int_A^B \hat{f}(\xi) e^{2\pi i \xi x} d \xi = f \ast h(x)$ where $h(x) = e^{i \pi x (B-A)}\frac{\sin(\pi x (B-A)/2)}{\pi x}$ is an ideal band-pass filter – 2017-02-25
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    Ah I see and the indefinite integral would be $g(A, x) = \int_0^A \hat{f}(\xi) e^{2\pi i \xi x)} d \xi$, so it can be interpreted as a band-passed version of the function? Thanks again! I might suggest moving some of the information in these comments into the main body of the answer. – 2017-02-25
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    @KFox The filter is said ideal because its impulse response $\frac{\sin x}{x}$ doesn't have a finite support and it isn't even causal, so in real life we can only compute some approximations of it – 2017-02-25