If $f \in L^1(-\infty, \infty)$ , and $G(u) = \frac{f(x+u) - f(x^+)}{\pi u}$, is $G(u) \in L^1(0^+, \infty)$ true?

Question

To be more detailed, if function $f(x)$ satisfies $\int_{-\infty}^{\infty}|f(x)|dx < \infty$ and assume that $G(u) = \frac{f(x+u) - f(x^+)}{\pi u}$, is it true that $lim_{K \rightarrow \infty} \int_{0^+}^{\infty}G(u) \cdot sin(Ku) \cdot du = 0$?

I got to this question by reading this tutorial on the proof of Fourier Integral Theorem: https://people.math.osu.edu/gerlach.1/math/BVtypset/node31.html

The proof is actually nice where most parts are clear. However I was stuck by some last steps which I summarized and put in this description. Please let me know if there's any error or misinterpretation of the original proof found.

EDITED:

Sorry for any confusion in the description, the $x^+$ notion means "one-sided limit to the right of $x$" and by $G(u) = \frac{f(x+u) - f(x^+)}{\pi u}$ I meant that $\frac{du}{dx} = 0$.

Answer 1

This is not true in general.

As a counter example, look at $f : t \mapsto exp(-t^2) \in L^1(\mathbb{R})$.

$\forall x \in \mathbb{R} , G(u) = \frac{exp\left(-(x+u)^2\right)-exp(-x^2)}{ u \pi} \notin L^1(\mathbb{R}^{+})$.

Therefore I have no idea why the Riemann-Lebesgue lemma can be applied here.