= Solution
Remove the two reciprocal tails by setting
$$
r(x)=w(x)-c_+u(x)+c_-u(-x).
$$
The assumed $O(x^{-2})$ remainder at each end and <local integrability> on bounded intervals imply $r\in L^1(\mathbb R)$. Its <Fourier transform> $\widehat r$ is therefore <continuous> at zero. Since $\widehat{u(-\mathord\cdot)}(\lambda)=\widehat u(-\lambda)$,
$$
\widehat w(\lambda)
=c_+\widehat u(\lambda)-c_-\widehat u(-\lambda)+\widehat r(\lambda).
$$
Parts (iv) and (v) show that both requested one-sided limits exist. If $L_+$ denotes the limit from positive frequencies and $L_-$ the limit from negative frequencies, then the common value $\widehat r(0)$ cancels and
$$
\begin{aligned}
L_+-L_-
&=(c_++c_-)
\lim_{\lambda\downarrow0}
\bigl[f_+(\lambda)-f_-(-\lambda)\bigr]\\
&=\boxed{-i\pi(c_++c_-)}.
\end{aligned}
$$
This is the <Fourier transform of a function with reciprocal tails>[universal jump caused by reciprocal tails].
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