# Fourier transform and distibution beloning to S'

I need prove, that a distribution $$\langle F_f,\phi \rangle= p.v. \int\limits_{-1/2}^{1/2}\frac{\phi(t)}{t\cdot \ln{|t|}}\mathrm{d}t$$ belongs to $S^\prime$ (adjoint to Schwartz space) and I need find, Fourier transform of the distribution.

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Is $F_f$ supposed to be the Fourier Transform of $f$? How is $f$ incorporated on the right-hand side? – robjohn May 23 '12 at 18:14
$f~$ is generating function for $F_f~$, so $$f=\frac{1}{t\cdot \ln{|t|}}$$ – user31497 May 23 '12 at 18:45
So are you looking for $$\int_{-1/2}^{1/2}\frac{e^{-2\pi i\xi t}}{t\log|t|}\,\mathrm{d}t$$ ? – robjohn May 23 '12 at 18:57
Yes, but I think there are some trick in this situation. And I can't prove, that $F_f\in S~'$. It's a main problem. – user31497 May 23 '12 at 19:18
No, every distribution is infinitely differentiable (in the weak sense). – Vobo May 24 '12 at 7:20

If $\phi\in\mathcal{S}$, $\psi(t)=\frac{\phi(t)-\phi(0)}{t}\in C^\infty$, and by the Mean Value Theorem, $\|\psi\|_{L^\infty}\le\|\phi'\|_{L^\infty}$ Thus, \begin{align} \lim_{\epsilon\to0}\left(\int_{-1/2}^{-\epsilon}\frac{\phi(t)}{t\log|t|}\,\mathrm{d}t+\int_{\epsilon}^{1/2}\frac{\phi(t)}{t\log|t|}\,\mathrm{d}t\right) &=\lim_{\epsilon\to0}\left(\int_{-1/2}^{-\epsilon}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t+\int_{\epsilon}^{1/2}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t\right)\\ &=\int_{-1/2}^{1/2}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t\tag{1} \end{align} Therefore \begin{align} \left|\text{p.v.}\int_{-1/2}^{1/2}\frac{\phi(t)}{t\log|t|}\,\mathrm{d}t\right|&=\left|\int_{-1/2}^{1/2}\frac{\psi(t)}{\log|t|}\,\mathrm{d}t\right|\\ &\le\left|\int_{-1/2}^{1/2}\frac{1}{\log|t|}\,\mathrm{d}t\right|\;\|\psi\|_{L^\infty}\\ &\le\frac{1}{\log(2)}\|\phi'\|_{L^\infty}\tag{2} \end{align} Inequality $(2)$ says that $\langle F_f,\cdot\rangle\in\mathcal{S}'$.
The Fourier Transform of the distribution would be \begin{align} \text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-2\pi i\xi t}}{t\log|t|}\,\mathrm{d}t &=-i\int_{-1/2}^{1/2}\frac{\sin(2\pi\xi t)}{t\log|t|}\,\mathrm{d}t\\ &=-2i\int_0^{1/2}\frac{\sin(2\pi\xi t)}{t\log|t|}\,\mathrm{d}t\tag{3} \end{align} I can not find a closed form for $(3)$.
@user31497: I think what you are asking is that since the Fourier transform of an $L^1$ function is bounded, how do we justify that $\frac{1}{x\log|x|}$ blows up near $0$. The Fourier transform of most $L^1$ functions are not themselves $L^1$. This means that the inverse Fourier transform does not necessarily converge absolutely. However, they converge weakly, that is $$\int f(x)\overline{\phi(x)}\,\mathrm{d}x=\int \hat{f}(x)\overline{\hat{\phi}(x)}\,\mathrm{d}x$$ for any $\phi\in\mathcal(S)$. – robjohn May 24 '12 at 0:00
Excuse me, but how did you find the Fourier Transform for this function? Do we have to use statement:$<\widehat{f},g>=<f,\widehat{g}>$? – user31497 May 24 '12 at 8:33
Yes. Plug in $\hat{g}(\tau) = \int \exp (-2\pi i \tau t) g(t) \mathrm{d}t$ into $\langle \mathcal{F}_f,\hat{g}\rangle$ from the definition and then interchange the order of integration. – Willie Wong May 24 '12 at 8:43