I need to prove that there is no odd function on $L^1$ whose Fourier transform is a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $g(\xi)=1/\log(\xi)$ for $\xi\geq 2$.
I am suggested to show that for every $0<\epsilon < t$ we have $$ \left| \int_\epsilon^t \frac{\sin(\xi)}{\xi} \; d\xi \right| \leq 4. $$
Having this, I had succesfully prove that such $g$ is not the Fourier transform of a $L^1$ function. I only need to prove the above inequality.
I proved that $$ \int_\epsilon^t \frac{\sin(\xi)}{\xi} \; d\xi = \int_0^{\infty} \frac{e^{-\epsilon y}(y\sin(\epsilon)+\cos(\epsilon))-e^{-ty}(y\sin(t)+\cos(t))}{1+y^2}\; dy $$ and I believe I can bound the latter integral, but I don't know how to proceed. Maybe I'm wrong and I chose the wrong way. Any help will be appreciated.
Since $0 \le \sin x \le x, 0 \le x \le 1, \left| \int_0^1 \frac{\sin(\xi)}{\xi} \; d\xi \right| \leq 1$ hence $\left| \int_\epsilon^\delta \frac{\sin(\xi)}{\xi} \; d\xi \right| \leq 1, 0 \le \epsilon \le \delta \le 1$.
Applying the second mean value theorem for $1 \le \epsilon \le t, \left| \int_\epsilon^t \frac{\sin(\xi)}{\xi} \; d\xi \right|=\frac{1}{\epsilon}\left| \int_\epsilon^\delta \sin(\xi) \; d\xi \right| \le 2, 1\le \epsilon \le \delta \le t$
Putting things together, if $0< \epsilon < t \le 1$, the integral is at most $1$, if $1 \le \epsilon < t$, the integral is at most $2$ and if $0 \le \epsilon \le 1 < t$, splitting at $1$ we get that the integral is at most $3$