prove if f(x), xf(x) ∈ L1(R) then integral of f(x)sin(wx)dw is defined, continuous, and differentiable at every point w

Question

if f(x), xf(x) ∈ L1(R) then F(w) = ∫f(x)sin(wx)dw is defined, continuous, and differentiable at every point w ∈ R.

Answer 1

Call $$F(w)=\int_{\mathbb{R}}f(x)\sin(wx)$$

Since $\int_{\mathbb{R}} |f(x)\sin(wx)|\leq\int_{\mathbb{R}}|f(x)|<\infty$, then $F(w)$ exists.

Now, let $h_n\to0$. Then

$$\begin{align}\frac{F(w+h_n)-F(w)}{h_n}&=\frac{\int_{\mathbb{R}} f(x)\sin((w+h_n)x)-\int_{\mathbb{R}} f(x)\sin((w)x)}{h_n}\\&=\int_{\mathbb{R}}f(x)\frac{\sin((w+h_n)x)-\sin(wx)}{h_n}\\&=\int_{\mathbb{R}}f(x)x\cos(z_nx)\end{align}$$ for some $w_n$ in the interval from $w$ to $w+h_n$.

Pointwise $f(x)x\cos(w_nx)\to xf(x)\cos(wx)$. Take into account that since $h_n\to0$, then $w_n\to w$ Since $|xf(x)\cos(z_nx)|\leq |xf(x)|\in L^1$ by the dominated convergence theorem $$\int_{\mathbb{R}}f(x)x\cos(z_nx)\to \int_{\mathbb{R}}xf(x)\cos(wx)$$

Since $h_n$ was an arbitrary sequence tending to $0$, this implies that $\lim_{h\to0}\frac{F(w+h)-F(w)}{h}$ exists for each $w$.

Therefore, $F(w)=\int_{\mathbb{R}}f(x)\sin(wx)$ is differentiable, and also continuous.