for which values of $x,\,$ $I_x = \int_{0}^{+\infty} \frac{(u\cos u - sin u)\cos ux}{u^3} \, du = 0$?

Question

I could prove that for $x = 1$ it's $0$

in fact if you compute the fourier transform of $f(x) =(1-x^2) \chi_{[-1,1]}$ you get $\hat{f}(u) =4\frac{\sin u - u\cos u}{u^3}$

on the one hand we have : $$I_1 = \frac{1}{2}\int_{\mathbb{R}} \frac{(u\cos u - sin u)\cos u}{u^3} \, du$$

and on the other hand we have $ \mathcal{F}[\frac{\delta(x-1)+\delta(x+1)}{2}] = \frac{1}{2} [e^{iu}+e^{-iu}] = \cos u$

then by Plancherel theorem : $I_1 = K \int_{\mathbb{R}} \frac{\delta(x-1)+\delta(x+1)}{2}(1-x^2) \chi_{[-1,1]}dx = \frac{K}{2}[0+0] = 0$

any ideas which other values of $x$ may give the same result ?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mc{I} \equiv {1 \over 2}\int_{\mathbb{R}}{\bracks{u\cos\pars{u} - \sin\pars{u}}\cos\pars{ux} \over u^{3}}\,\dd u:\ {\Large ?}}$.

\begin{align} \mc{I} & \equiv {1 \over 2}\int_{\mathbb{R}}{\bracks{u\cos\pars{u} - \sin\pars{u}}\cos\pars{ux} \over u^{3}}\,\dd u = {1 \over 2}\int_{-\infty}^{\infty}\bracks{-\int_{0}^{1}t\,\sin\pars{ut}\,\dd t}{\cos\pars{u\verts{x}} \over u}\,\dd u \\[5mm] & = -\,{1 \over 4}\int_{0}^{1}t\int_{-\infty}^{\infty} {\sin\pars{\bracks{t + \verts{x}}u} + \sin\pars{\bracks{t - \verts{x}}u} \over u}\,\dd u\,\dd t \\[5mm] & = -\,{1 \over 4}\int_{0}^{1}t\bracks{% \mrm{sgn}\pars{t + \verts{x}}\pi + \mrm{sgn}\pars{t - \verts{x}}\pi}\,\dd t \\[5mm] &= -\,{1 \over 4}\,\pi\bracks{% {1 \over 2} +\int_{0}^{1}t\,\,\mrm{sgn}\pars{t - \verts{x}}\,\dd t} \end{align}

---

Then, \begin{align} \mc{I} & \equiv {1 \over 2}\int_{\mathbb{R}}{\bracks{u\cos\pars{u} - \sin\pars{u}}\cos\pars{ux} \over u^{3}}\,\dd u \\[5mm] & = -\,{1 \over 4}\,\pi\bracks{% {1 \over 2} + \int_{0}^{1}\bracks{t > \verts{x} }t\,\dd t - \int_{0}^{1}\bracks{t < \verts{x}}t\,\dd t} \\[5mm] & = -\,{1 \over 4}\,\pi\bracks{% {1 \over 2} + \bracks{0 < \verts{x} < 1}\int_{\verts{x}}^{1}t\,\dd t - \bracks{0 < \verts{x} < 1}\int_{0}^{\verts{x}}t\,\dd t - \bracks{\verts{x} > 1}\int_{0}^{1}t\,\dd t} \\[5mm] & = -\,{1 \over 4}\,\pi\bracks{% {1 \over 2} + \bracks{0 < \verts{x} < 1}\pars{{1 \over 2} - {1 \over 2}\,x^{2}} - \bracks{0 < \verts{x} < 1}{1 \over 2}\,x^{2} - \bracks{\verts{x} > 1} {1 \over 2}} \\[5mm] & = -\,{1 \over 4}\,\pi\bracks{% {1 \over 2} + \bracks{0 < \verts{x} < 1}\pars{{1 \over 2} - x^{2}} - \bracks{\verts{x} > 1}{1 \over 2}} \\[5mm] & = \bbx{{1 \over 4}\,\pi\bracks{0 < \verts{x} < 1}\pars{x^{2} - 1}} \end{align}