= Solution
For smooth rapidly decaying initial data, apply the <Fourier transform> in $x$ and differentiate under the <integral>. The <Fourier transform of a derivative> turns the <wave equation> into
$$
\widetilde v_{tt}+k^2\widetilde v=0,\quad\widetilde v(k,0)=\widetilde f(k),\quad\widetilde v_t(k,0)=\widetilde g(k).
$$
Solving this constant-coefficient <ordinary differential equation> gives
$$
\boxed{\widetilde v(k,t)=\widetilde f(k)\cos kt+\widetilde g(k)\frac{\sin kt}{k}}.
$$
At $k=0$ the quotient means its continuous limit $t$, so $\widetilde v(0,t)=\widetilde f(0)+t\widetilde g(0)$.
Since $\cos kt=(e^{ikt}+e^{-ikt})/2$, the <Translation property of the Fourier transform> inverts the first term as $(f(x+t)+f(x-t))/2$. Also $K_t(x)=\tfrac12\mathbf1_{[-t,t]}(x)$ has <Fourier transform> $\sin kt/k$, including its limiting value at zero. The <convolution theorem> therefore inverts the second term as
$$
(K_t*g)(x)=\frac12\int_{-t}^t g(x-s)\,ds=\frac12\int_{x-t}^{x+t}g(\xi)\,d\xi.
$$
This proves the <D'Alembert formula>
$$
\boxed{v(x,t)=\frac{f(x-t)+f(x+t)}2+\frac12\int_{x-t}^{x+t}g(\xi)\,d\xi}.
$$
The formula itself extends beyond the rapidly decaying class: $f\in C^2$ and $g\in C^1$ suffice for a classical solution, verified by differentiation and the two initial conditions. The decay assumption above justifies the ordinary transform derivation without distributional qualifications.
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