Goal: Show that
$$ [-2 \pi i x f(x)]^{\wedge}(\xi) = {d \over d\xi} \widehat{f}(\xi) $$
where $\wedge$ denotes the operation of the Fourier Transform:
$$ \widehat{f}(\xi) = \int_{- \infty}^{\infty} e^{- 2 \pi i \xi x} f(x) dx $$
Proof from Book (with problematic portions highlighted):
Problem: It is not apparent to me why the three highlighted portions of this proof are true. That is, it is not apparent to me why, for $|x| > N$ and $h$ small enough, that each line of this inequality holds:
\begin{align*} \left| [\widehat{f}(\xi + h) - \widehat{f}(\xi)]/h - [-2\pi ixf(x)]^{\wedge}(\xi) \right| &= \left| \int_{\mathbb{R}} f(x) e^{-2 \pi i x \xi} \left[ {e^{- 2 \pi i x h} - 1\over h} + 2 \pi i x\right] dx \right| \\ &\le 2\pi \int_{|x|> N} |x| |f(x)| \left| {e^{- 2 \pi i x h} - 1 \over 2 \pi x h} + i \right| dx \end{align*}
I think this may work easier for you instead: $$ \frac{e^{-2\pi ixh}-1}{h}+2\pi ix = -2\pi i x\left[\frac{1}{h}\int_{0}^{h}(e^{-2\pi ixt}-1)dt\right] \\ = -2\pi ix \left[\frac{1}{h}\int_{0}^{h}\int_{0}^{t}(-2\pi ix)e^{-2\pi ixs}dsdt\right] \\ = 4\pi^2 x^2\frac{1}{h}\int_{0}^{h}\int_{0}^{t}e^{-2\pi ixs}dsdt. $$