= Solution
Let
$$
v(\lambda)=-i\int_1^\infty\frac{e^{-i\lambda x}}{x^2}\,dx.
$$
The integral is <absolute convergence>[absolutely convergent]. For a <Schwartz function> $\varphi$, <Fubini's theorem> and <integration by parts> in $\lambda$ give
$$
\begin{aligned}
\int_{\mathbb R}\varphi'(\lambda)v(\lambda)\,d\lambda
&=-i\int_1^\infty\frac1{x^2}
\left(\int_{\mathbb R}\varphi'(\lambda)e^{-i\lambda x}\,d\lambda\right)dx\\
&=-\int_1^\infty\frac1x
\left(\int_{\mathbb R}\varphi(\lambda)e^{-i\lambda x}\,d\lambda\right)dx\\
&=\langle\widehat u,\varphi\rangle.
\end{aligned}
$$
Equivalently, $v'=-\widehat u$ as a <distributional derivative>, and hence
$$
\boxed{\langle\widehat u,\varphi\rangle
=\int_{\mathbb R}\varphi'(\lambda)v(\lambda)\,d\lambda}.
$$
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