Identity for the space-time fourier transform of the "forced term" in the airy equation

Question

Let's consider a function $f(t,x)$ smooth and sufficiently fast decaying in $\mathbb{R}^2$, where $t,x\in\mathbb{R}$. Consider also the time-dependent operator $V(t)$ given by $$ V(t)g:=\int_{-\infty}^{+\infty} e^{i(x\xi+t\xi^3)}\mathcal{F}_x[g](\xi)d\xi,\quad \hbox{with} \quad g:\mathbb{R}\to\mathbb{R} $$ where $\mathcal{F}_x[g]$ denotes the spatial-Fourier transform of the function $g(x)$. The operator $V(t)$ is usually called the Airy group. Now, I would like to prove the following identity: for any $t>0$, we have: $$ \int_0^tV(t-t')f(t',x)dt'=\int_{\mathbb{R}}\int_{\mathbb{R}}e^{ix\xi}\dfrac{e^{it\tau}-e^{it\xi^3}}{\tau-\xi^3}\mathcal{F}_{t,x}[f](\tau,\xi)d\xi d\tau, $$ where $\mathcal{F}_{t,x}[f]$ denotes the space-time Fourier transform of $f(t,x)$. My thoughts were trying to write the integrand $V(t-t')f(t',x)$ as the Fourier transform of the anti-Fourier transform, and then switching integrals and doing things like that, but I couldn't succeed, mainly because I am not sure how to deal with the time-Fourier transform and the integral on $(0,t)$. Does anyone has any hint?

Answer 1

Just for the record, I think I succeeded computing the solution. First, by writing $$ V(t-t')f(t',x)=\int_{\mathbb{R}}\int_{\mathbb{R}}e^{ix\xi}e^{it'\tau}e^{i(t-t')\xi^3}\mathcal{F}_{t',x}[f](\tau,\xi)d\tau d\xi. $$ Then, notice that $$ \int_0^te^{i(\tau-\xi^3)t'}dt'=\dfrac{i(1-e^{i(\tau-\xi^3)t})}{\tau-\xi^3}. $$ Hence, by using the previous two identities we conclude $$ \int_0^tV(t-t')f(t',x)dt'=i\int_{\mathbb{R}}\int_{\mathbb{R}}e^{ix\xi}\Big(\dfrac{e^{it\xi^3}-e^{it\tau}}{\tau-\xi^3}\Big)\mathcal{F}_{t,x}[f](\tau,\xi)d\tau d\xi. $$ I think this should be correct. However, I feel like I am missing a "$i$" factor multiplying the whole expression, in order to recover exactly the identity the book claim. Anyways, usually the book don't care about constants, so I guess this is the exact result. If anyone see any typo I would really appreciate.