d'Alembert's solution to the wave equation via Fourier Transforms

I am trying to solve the wave equation $$v_{tt} = v_{xx} \text{ on } (x,t) = (-\infty, \infty) \times (0,\infty)$$ with initial conditions $$v(x,0) = f(x), \quad v_t (x,0) = g(x)$$ I have shown that with respect to $$x$$, the Fourier transform of $$v$$ satisfies $$\hat v(k,t) = \hat f(k)\cos kt + \frac{\hat g(k)}{k} \sin kt$$ By inverse Fourier transform, $$v(x,t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\hat v(k,t) e^{ikx}dk$$ Applying the integral to the first term and using the exponential definition of cos, we have $$\frac{1}{2}\left(f(x-t) + f(x+t)\right)$$ However, I am unsure how to show that $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\hat g(k)}{k}\sin (kt) e^{ikx}dk = \frac{1}{2}\int_{x-t}^{x+t}g(\xi)d\xi$$ I assume it has something to do with the Heaviside function since the Fourier transform of $$\sin(kt)/k$$ is proportional to the Heaviside function, but the factor of $$\hat g(k)$$ is making things awkward. Any help would be greatly appreciated.

\begin{align} \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\hat g(k)}{k}\sin (kt) e^{ikx}dk &=\frac{1}{4\pi i}\int_{-\infty}^{\infty}\frac{\hat g(k)}{k}\left[e^{ik(x+t)}-e^{ik(x-t)}\right]dk \\ &=\frac{1}{4\pi}\int_{-\infty}^{\infty}\hat g(k)\int_{x-t}^{x+t}e^{ik\xi}\,d\xi\, dk \\ &=\frac{1}{2}\int_{x-t}^{x+t}\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat g(k)e^{ik\xi}\,dk\,d\xi \\ &=\frac{1}{2}\int_{x-t}^{x+t}g(\xi)\,d\xi \end{align}