Function of $C_0(\mathbb{R})$ I need to prove that 
$$g(x) = \text{p.v.} \int\limits_{-1/2}^{1/2}\frac{e^{-itx}}{t\cdot \ln{|t|}}dt
$$ 
is function of $C_{0}(\mathbb{R})$. 
So, I need to prove that 
$$
\lim\limits_{|x|\to\infty}g(x)=0,
$$ 
but I don't know how. Please help me.  
 A: Continuing from where I left off in answer to another question, because $\frac{1}{t\log|t|}$ is odd,
$$
\begin{align}
\text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-itx}}{t\log|t|}\,\mathrm{d}t
&=-i\int_{-1/2}^{1/2}\frac{\sin(tx)}{t\log|t|}\,\mathrm{d}t\\
&=-2i\int_0^{1/2}\frac{\sin(tx)}{t\log|t|}\,\mathrm{d}t\\
&=-2i\int_0^{1/2}\frac{1}{t\log|t|}\,\mathrm{d}\frac{1-\cos(tx)}{x}\\
&=4i\frac{1-\cos(x/2)}{\log(2)x}+2i\int_0^{1/2}\frac{1-\cos(tx)}{x}\,\mathrm{d}\frac{1}{t\log|t|}\\
&=4i\frac{1-\cos(x/2)}{\log(2)x}-2i\int_0^{1/2}\frac{1-\cos(tx)}{x}\frac{1+\log(t)}{(t\log(t))^2}\,\mathrm{d}t\\
&=4i\frac{1-\cos(x/2)}{\log(2)x}-\frac{2i}{x}\int_0^{1/2}\frac{1-\cos(tx)}{t^2}\frac{1+\log(t)}{\log(t)^2}\,\mathrm{d}t\tag{1}
\end{align}
$$
We have that $(1)$ is odd and for $x>4$,
$$
\left|4i\frac{1-\cos(x/2)}{\log(2)x}\right|\le\frac{8}{\log(2)x}\tag{2}
$$
and
$$
\begin{align}
&\left|\frac{2i}{x}\int_0^{1/2}\frac{1-\cos(tx)}{t^2}\frac{1+\log(t)}{\log(t)^2}\,\mathrm{d}t\right|\\
&\le\frac2x\left(\int_0^{1/x}\frac{x^2}{2}\frac{1}{\log(x)}\mathrm{d}t+\int_{1/x}^{1/\sqrt{x}}\frac{2}{t^2}\frac{1}{\log(\sqrt{x})}\mathrm{d}t+\int_{1/\sqrt{x}}^{1/2}\frac{2}{t^2}\frac{1}{\log(2)}\mathrm{d}t\right)\\
&\le\frac2x\left(\frac{x}{2\log(x)}+\frac{4x}{\log(x)}+\frac{2\sqrt{x}}{\log(2)}\right)\\
&=\frac{9}{\log(x)}+\frac{4}{\log(2)\sqrt{x}}\tag{3}
\end{align}
$$
Thus, for $|x|>4$,
$$
\left|\text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-itx}}{t\log|t|}\,\mathrm{d}t\right|\le\frac{9}{\log|x|}+\frac{4}{\log(2)\sqrt{|x|}}+\frac{8}{\log(2)|x|}\tag{4}
$$
So that
$$
\lim_{|x|\to\infty}\text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-itx}}{t\log|t|}\,\mathrm{d}t=0\tag{5}
$$
