Let f be $L^1(R)$ and odd function. Then, for any positive $a < A$, there is $M>0$ such that $$ \left|{\int_{a}^A \frac{ \hat{f}(\alpha)}{\alpha} \ d\alpha}\right| \leq M $$
($\hat{f}$ is the fourier transform of transform of f)
holds. How can I show that such M exists?
If $f$ is odd and integrable, then $$ g(x) = \int_{0}^{x}f(y)dy $$ is even and bounded with limits at $\pm\infty$. Therefore, \begin{align} \int_{-R}^{R}e^{-isx}f(x)dx & = e^{-isx}g(x)|_{x=-R}^{R}+is\int_{-R}^{R}e^{-isx}g(x)dx \\ & = -2i\sin(sR)g(R)+is\int_{-R}^{R}e^{-isx}g(x)dx. \\ \frac{1}{2is}\int_{-R}^{R}e^{-isx}f(x)dx & = -\frac{\sin(sR)}{s}g(R)+\int_{0}^{R}\cos(sx)g(x)dx. \end{align} Integrate both sides over $[a,A]$ assuming $0\notin [a,A]$: $$ \int_{a}^{A}\frac{1}{2is}\int_{-R}^{R}e^{-isx}f(x)dxds \\ = -\int_{a}^{A}\frac{\sin(sR)}{s}dsg(R)+\int_{0}^{R}\frac{\sin(Ax)-\sin(ax)}{x} g(x)dx \\ = -\int_{aR}^{AR}\frac{\sin(u)}{u}du g(R)+\int_{0}^{R}\frac{\sin(Ax)-\sin(ax)}{x} g(x)dx $$ Now suppose that $0 \notin [a,A]$. The integral on the left converges as $R\rightarrow\infty$ because the integrand converges uniformly to $\sqrt{2\pi}\hat{f}(s)/2is$. The following converges to $0$ as $R\rightarrow\infty$: $$ \int_{aR}^{AR}\frac{\sin(u)}{u}du g(R) $$ This is because $\int_{0}^{x}\frac{\sin(u)}{u}du$ is uniformly bounded in $x$ and converges to $\pm\frac{\pi}{2}$ as $x\rightarrow \pm\infty$, and $g(x)$ is uniformly bounded. Therefore, the remaining integral on the right must also converge as $R\rightarrow\infty$. Hence, if $0\notin[a,A]$, $$ \sqrt{2\pi}\int_{a}^{A}\frac{1}{2is}\hat{f}(s)ds = \int_{0}^{\infty}\frac{\sin(Ax)-\sin(ax)}{x}g(x)dx. $$ The right side is guaranteed to exist as an improper integral. Integrating the right side by parts gives $$ \left.\int_{0}^{x}\frac{\sin(Ay)-\sin(ay)}{y}dyg(x)\right|_{x=0}^{\infty} - \int_{0}^{\infty}\int_{0}^{x}\frac{\sin(Ay)-\sin(ay)}{y}dy f(x)dx \\ = -\int_{0}^{\infty}\int_{0}^{x}\frac{\sin(Ay)-\sin(ay)}{y}dy f(x)dx $$ which is uniformly bounded by $$ 2\int_{0}^{\pi/2}\frac{\sin(y)}{y}dy\int_{0}^{\infty}|f(x)|dx = \int_{0}^{\pi/2}\frac{\sin(y)}{y}dy\int_{-\infty}^{\infty}|f(x)|dx. $$ So it appears to me that $$ \left|\int_{a}^{A}\frac{\hat{f}(s)}{s}ds\right|\le\sqrt{\frac{2}{\pi}}\int_{0}^{\pi/2}\frac{\sin(x)}{x}dx\int_{-\infty}^{\infty}|f(x)|dx. $$ So I think the best constant $M$ is $$ M = \sqrt{\frac{2}{\pi}}\int_{0}^{\pi/2}\frac{\sin(x)}{x}dx\int_{-\infty}^{\infty}|f(x)|dx $$
