= Solution
Expand in the complete <Fourier sine series> $\widetilde y(\omega,x)=\sum_{n\geq1}b_n(\omega)\sin(k_nx)$ and apply <Fourier inversion>. Substitution of part (ii) gives
$$
\boxed{y(t,x)=\frac1{\pi L}\sum_{n=1}^\infty\sin(k_nx_0)\sin(k_nx)\int_{-\infty}^\infty\frac{\widetilde F(\omega)e^{i\omega t}}{k_n^2-\omega^2}\,d\omega,\qquad C=\frac1{\pi L}.}
$$
To check the source normalization, apply $\partial_t^2-\partial_x^2$ mode by mode. It cancels the denominator; the remaining inverse transform gives $2\pi F(t)$, and completeness gives
$$
\frac2L\sum_{n\geq1}\sin(k_nx_0)\sin(k_nx)=\delta(x-x_0).
$$
The endpoints vanish termwise. These identities establish the equation in the distributional sense, with classical validity wherever the series and derivatives converge suitably.
The displayed real-axis integrals in the paper are not ordinary convergent integrals at their <poles>. One definite choice is the <Cauchy principal value> at each $\omega=\pm k_n$. If a causal solution is wanted, use $[k_n^2-(\omega-i0)^2]^{-1}$ instead. Its inverse transform is $H(t)\sin(k_nt)/k_n$, so the causal solution is
$$
y_{\rm ret}(t,x)=\frac2L\sum_{n\geq1}\frac{\sin(k_nx_0)\sin(k_nx)}{k_n}\int_{-\infty}^tF(s)\sin[k_n(t-s)]\,ds,
$$
when the past integral exists. The principal-value choice instead uses the time-symmetric kernel $\sin(k_n|t|)/(2k_n)$. No temporal initial data are given, so homogeneous sine and cosine modes remain permissible additions. This is the <causal Fourier resolvent of a point-driven fixed string>.
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