Let $f$ be piecewise-smooth and continuous on $\mathbb{R}$ such that $f,f'\in L^2(\mathbb{R})$. I want to show that
$$\int_{-\infty}^{\infty} (1+\vert x\vert^2)\vert \hat{f}(x)\vert^2 dx<\infty$$
Where $\hat{f}$ is the Fourier transform of $f$. Since $f,f'\in L^2(\mathbb{R})$, we have $\Vert f\Vert_{L^2} = \Vert \hat{f}\Vert_{L^2}$ and $\Vert f'\Vert_{L^2} = \Vert \hat{f'}\Vert_{L^2}$. Thus
\begin{align*} \int_{-\infty}^{\infty} (1+\vert x\vert^2)\vert \hat{f}(x)\vert^2 dx &= \int_{-\infty}^{\infty} \vert \hat{f}(x)\vert^2 dx + \int_{-\infty}^{\infty} \vert x\hat{f}(x)\vert^2 dx\\ &= \Vert \hat{f}\Vert_{L^2} + \int_{-\infty}^{\infty} \vert x\hat{f}(x)\vert^2 dx\\ &= \Vert f\Vert_{L^2} + \int_{-\infty}^{\infty} \vert x\hat{f}(x)\vert^2 dx.\\ \end{align*}
Now how do I evaluate the other integral? I'm sure I have to use $\Vert f'\Vert_{L^2} = \Vert \hat{f'}\Vert_{L^2}$ but i have no idea how. Any hints? I've been stuck for days now. Thanks!
Integrate by part to get $$\hat {f'}(\xi)=\int_{\mathbb R}f'(x)e^{-2\pi i\xi x}dx=\underbrace{\left.f(x)e^{2\pi i\xi x}\right|_{-\infty}^{\infty}}_{=0}+2\pi i\xi\int_{\mathbb R}f(x)e^{-2\pi i\xi x}dx=2\pi i\xi\hat f(\xi)$$ therefore $\int_{\mathbb R}|\xi\hat {f'}(\xi)|^2d\xi$ is, up to a constant, $\|f'\|_{L^2}^2$