Consider $u \in \mathcal{S}'(\mathbb{R})$ where $$u(x) = \frac{xH(-x)}{e^x + 1}, \quad x \in \mathbb{R},$$ where $H$ is the Heaviside step function. The goal is to evaluate $$\lim_{\xi \rightarrow 0} \xi^2 \hat{u}(\xi),$$ where $\hat{u}$ represents the Fourier Transform of ${u}$.
Intuitively, for $x \rightarrow -\infty$, we have $u \rightarrow x$, and we should expect some sort of $\delta'$ to appear, which upon multiplication with $\xi^2$ and taking the limit as $\xi$ goes to $0$, we have that the $\delta'$ term does not contribute to the integral. However, Mathematica gives that the Fourier Transform is given by $1$, which comes from the fact that $\hat{u}$ has a term in $\frac{1}{\xi^2}$. Thus, I would like to ask if anyone has any idea as to how we can compute this integral (or at least the required limits)? (My guess was that it has something to do with contour integrals, but I could be wrong about this.) Thanks!
Convention: We take the Fourier Transform $\hat{u}$ to be given by $$\hat{u}(\xi) = \int_{\mathbb{R}} u(x)e^{-ix\xi} \mathrm{d}x.$$
One useful Fourier transform trick is that
$$\hat{f}(0) = \int_{\Bbb{R}}f(x)e^{-ix\cdot0}\:dx = \int_{\Bbb{R}}f(x)\:dx$$
Another Fourier transform property tells us that
$$\xi^2\hat{u}(\xi) \leftrightarrow -u''(x)$$
Therefore the quantity we want is
$$\lim_{\xi\to0}\xi^2\hat{u}(\xi) = \int_{\Bbb{R}}-u''(x)\:dx = u'(-\infty)-u'(\infty)$$
Calculating the derivative we have
$$\left(\frac{xH(-x)}{e^x+1}\right)' = \frac{H(-x)}{e^x+1}-\frac{x\delta(x)}{e^x+1}-\frac{xe^xH(-x)}{(e^x+1)^2} = \frac{H(-x)}{e^x+1}-\frac{xe^xH(-x)}{(e^x+1)^2}$$
which means
$$\lim_{\xi\to0}\xi^2\hat{u}(\xi) = 1$$