What's wrong with this derivation of d'Alembert's formula?

Question

Problem is
$u_{t t}-a^2 u_{x x}=0$, $u(x,t=0)=g(x)$, and $u_t(x,t=0)=h(x)$ for $-\infty

If one use the Fourier transform, that is, make all the equations and initial conditions multiplied by a factor $ e^{-i\omega x}/{2\pi} $ and then make integral with $x$, the new problem is $U_{t t}+{(\omega a)}^2 U=0$, $U(\omega,t=0)=G(x)$, and $U_t(\omega,t=0)=H(x)$ for $-\infty<\omega<\infty$, with the Fourier transform of $g(x)$ is $G(\omega)$, that of $h(x)$ is $H(\omega)$.

The general solution of the new problem is $U(\omega,t)=A(\omega)e^{i\omega a t} + B(\omega)e^{-i\omega a t}$.

And form the initial conditions, one get
$A(\omega)=\frac{1}{2}G(\omega)+\frac{1}{2 a}H(\omega)/(ik)$, and $B(\omega)=\frac{1}{2}G(\omega)-\frac{1}{2 a}H(\omega)/(ik)$.

So $U(\omega,t)=\frac{1}{2}G(\omega)e^{i\omega at}+\frac{1}{2 a}H(\omega)/(ik)e^{i\omega at} + \frac{1}{2}G(\omega)e^{-i\omega at}-\frac{1}{2 a}H(\omega)/(ik)e^{-i\omega at}$.

Make the inverse Fourier transform,one get $u(x,t)=\frac{1}{2}[g(x+at)+g(x-at)]+\frac{1}{2a}\int _{x-at}^{x+at} h(x')d{x'}$, and it is J. R. d'Alembert's formula.

My question is how to get the inverse Fourier transform of $U(\omega,t)$. Could you help me?

The inverse Fourier transform of $H(\omega)/(ik)e^{i\omega at}$ is $\int _b^xh(x'+at)d{x'}$, and that of $H(\omega)/(ik)e^{-i\omega at}$ is $\int _b^xh(x'-at)d{x'}$, where $b$ is a constant.

But $\int _b^xh(x'+at)d{x'}-\int _b^xh(x'-at)d{x'}=\int _{x-at}^{x+at}h(x')d{x'}+\int _{b+at}^{b-at}h(x')d{x'} \neq {\int _{x-at}^{x+at}h(x')d{x'}}$.

What is wrong with my derivation? Any advice and help are welcome!