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I asked the following question on mathstack but didn't receive any answers. I suspect that this question has a simple answer but I haven't thought about Fourier transforms in a while so am being sluggish at figuring it out. Any help/comments/suggestions would be appreciated.

Consider a function $f$ that has smooth Fourier transform $\widehat{f}$ with compact support. In a particular problem I am considering, it would be useful to be able to compute the integral

$$\int_{\mathbb{R}} \widehat{f}(u) \ \frac{(e^{h-2\pi i u}-1)}{h-2\pi i u} \ du$$

where $h >0$. Is anything known about this integral? In particular can we compute this integral for general $f$ as above?

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1 Answer 1

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By Fourier inversion formula, $$ \int_{-\infty}^\infty e^{-2\pi i u x} \hat f(u) du = f(x) $$ for all $x$ (up to some constant factors that depend on normalisation of the Fourier transform). By Fubini's theorem, $$ \int_0^1 e^{h x} f(x) dx = \int_{-\infty}^\infty \frac{e^{h - 2\pi i u} - 1}{h - 2 \pi i u} \hat f(u) du . $$

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