I am having trouble writing a formal solution to the PDE $$u_{xx}+u_{tt}=g(x,t),$$ for $x\in (-\infty,\infty)$ and $t\in (-\infty,\infty)$.
If I take the Fourier transform of the PDE with respect to $x$ and $t$, I get that $$\hat{u}(\lambda,\mu)=-\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{g(x,t)}{\lambda^2+\mu^2}e^{-i(\lambda x+\mu t)} \ dx \ dt.$$ But I am having serious problems calculating the inverse Fourier transform of $\hat{u}(\lambda,\mu)$. I get that $$u(x,t)=-\frac{1}{4\pi^2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{g(x,t)}{\lambda^2+\mu^2}e^{-i(\lambda x+\mu t)+i(\lambda x+\mu t)} \ dx \ dt \ d\lambda \ d\mu,$$ which clearly doesn't make sense as the exponential term cancels. It is a bit frustrating as I'm having trouble building on from my knowledge of $1$D Fourier transforms.
It's wise to use dummy variables for the first integral, say
$$ \hat u(\lambda,\mu) = \frac{1}{2\pi}\iint_\limits{(s,\tau) \in \Bbb R^2} g(s,\tau) \frac{e^{-i(\lambda s + \mu \tau)}}{\lambda^2 + \mu^2} ds\ d\tau $$
The inverse transform of this is
\begin{align} u(x,t) &= \frac{1}{2\pi}\iint_\limits{(\lambda,\mu)\in\Bbb R^2} \hat u(\lambda,\mu)e^{+i(\lambda x + \mu t)} d\lambda d\mu \\ &= \frac{1}{2\pi}\iint_\limits{(s,t)\in\Bbb R^2}g(s,t) \left(\frac{1}{2\pi}\iint_\limits{(\lambda,\mu)\in\Bbb R^2} \frac{e^{+i\big(\lambda(x-s)+\mu(t-\tau)\big)}}{\lambda^2+\mu^2} d\lambda\ d\mu \right)ds\ d\tau \end{align}
Let $h(x,y)$ be the inverse Fourier transform of $\hat h(\lambda,\mu) = \dfrac{1}{\lambda^2+\mu^2}$, then the result is a convolution:
$$ u(x,y) = \frac{1}{2\pi}\iint_\limits{(s,\tau)\in\Bbb R^2} g(s,\tau) h(x-s,t-\tau) ds\ d\tau $$