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So I have the Fourier transform

$$ \widehat{f}(\omega)=\frac{1}{1+|\omega|} $$

of some function $f(x)$.

I need to know if the two integrals below converge or not. $$ \int_{-\infty}^{\infty}|xf(x)|dx < \infty $$

$$ \int_{-\infty}^{\infty}|f(x)|^{2}dx < \infty $$

I thing I should use Plancherel Theorem on the second one. But I'm not sure whether it meets the theorem conditions or not, because the theorem states that the integral $ \int_{-\infty}^{\infty}|\widehat{f}(\omega)|^{2}dx $ converges and not the integral over $|f(x)|^{2}$.

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  • $\begingroup$ I would be more worried with the first integral. The second is trivial after a correct application of Plancherel's identity: $$\int_\mathbb{R}\lvert f(x)\rvert^2\, dx=\int_\mathbb{R}\lvert \hat{f}(\omega)\rvert^2\, d\omega.$$(In particular, the integral in the left hand side is finite if and only if the integral in the right hand side is.) $\endgroup$ – Giuseppe Negro Sep 18 '15 at 21:48
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The Fourier tranform $\hat{f}(w)$ is in $L^{2}$. So the function $f$ is in $L^{2}$ as well. However, $$ \int_{-\infty}^{\infty}|xf(x)|dx = \infty. $$ To see why, \begin{align} -ixf(x)&=(-ix)L^{2}\mbox{-}\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{e^{iwx}}{1+|w|}dw \\ &= (-ix)L^{2}\mbox{-}\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{0}^{R}\frac{2i\sin(wx)}{1+w}dw \\ &= \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{x\sin(wx)}{1+w}dw \\ &= -\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{1}{1+w}\frac{d}{dw}\cos(wx)dw \\ &=-\frac{2}{\sqrt{2\pi}}\left[\left.\frac{1}{1+w}\cos(wx)\right|_{w=0}^{\infty}+\int_{0}^{\infty}\frac{\cos(wx)}{(1+w)^{2}}dw\right] \\ &=\frac{2}{\sqrt{2\pi}}-\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{\cos(wx)}{(1+w)^{2}}dw \end{align} Therefore, by the Riemann-Lebesgue lemma, $$ \lim_{x\rightarrow\infty}(-ix)f(x)=\sqrt{\frac{2}{\pi}}, $$ which prevents $xf(x)$ from being absolutely integrable. However $xf(x)$ is square integrable because $$ \frac{1}{(1+|w|)^{2}}\frac{w}{|w|}=\hat{f}'(w)=\widehat{(-ix)f(x)} $$ is square integrable.

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