I encounter something strange for me.
Usually for $z \in R$ and $t \in R$, $\int_{-\infty}^\infty dt e^{i (k-z)t}=2 \pi \delta(k-z)$. And then one has $$\int_{-\infty}^\infty dk \int_{-\infty}^\infty dt f(k) e^{i (k-z)t} =\int_{-\infty}^\infty dk f(k) 2 \pi \delta(k-z) =2 \pi f(z).$$
But I verified numerically that the above equation $$\int_{-\infty}^\infty dk \int_{-\infty}^\infty dt f(k) e^{i (k-z)t} =2 \pi f(z) $$ also hold for $z \in C$.
For example, for an randomly picked function $f(k)=k^7 e^{-k^2}$ and randomly picked complex number $z=1+2i$, one can verify numerically that $$\int_{-\infty}^\infty dk \int_{-\infty}^\infty dt f(k) e^{i (k-z)t} = -28943.8 - 20162.6 i =2 \pi f(z). $$
Why? How do one understand this?
I thought about whether one can still use the delta function argument as for $z \in R$. But here $\int_{-\infty}^\infty dt e^{i (k-z)t}$ is no longer proportional to the delta function $2 \pi \delta(k-z)$ since $z \in C$. Moreover, even if one can somehow treat it as a delta function, the integration path for $k$ is on the real axis and does not pass $z$.
Assume that $f\in\mathcal{S}(\mathbb{R}).$ For $\xi\in\mathbb{R}$ let $$ \hat{f}(\xi) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{-ix\xi} \, dx. $$
Then, for $x\in\mathbb{R}$, $$ f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi) e^{ix\xi} \, d\xi. $$
Now, for $x,y\in\mathbb{R},$ set $$ F(x,y) = \int \hat{f}(\xi) e^{i(x+iy)\xi} \, d\xi = \int \hat{f}(\xi) e^{-y\xi} e^{ix\xi} \, d\xi . $$ Obviously, $F(x,0)=f(x)$ but we want to show that, under some conditions, $F(x,y)=f(x+iy).$ Our first condition will be that $f$ is analytic. Then we want to show that $x+iy \mapsto F(x,y)$ is analytic.
Formally we have $$ \begin{align} \partial_y F(x,y) &= \int (-\xi) \hat{f}(\xi) e^{-y\xi} e^{ix\xi} \, d\xi\\ \partial_x F(x,y) &= \int (i\xi) \hat{f}(\xi) e^{-y\xi} e^{ix\xi} \, d\xi \\ \end{align} $$ so $i \partial_x F(x,y) = \partial_y F(x,y).$ It thus looks like $F$ satisfies the Cauchy-Riemann conditions. But this actually requires that the integrands $(-\xi) \hat{f}(\xi) e^{-y\xi}$ and $(i\xi)\hat{f}(\xi) e^{-y\xi}$ are both in $L^1(\mathbb{R}).$ Therefore, our second condition will be that $\xi \hat{f}(\xi) e^{-y\xi}$ is in $L^1(\mathbb{R})$ for all $y\in\mathbb{R}.$
Since both $f$ and $x+iy \mapsto F(x,y)$ are analytic, and $F(x,0)=f(x),$ we in fact have $F(x,y)=f(x+iy),$ i.e. $$ \int_{-\infty}^{\infty} \left( \frac{1}{2\pi} \int_{-\infty}^{\infty} f(t) e^{-it\xi} \, dt \right)\hat{f}(\xi) e^{i(x+iy)\xi} \, d\xi = f(x+iy). \label{analytic-fourier-inverse}\tag{*} $$
Your example function $f(x) = x^7 e^{-x^2}$ is analytic and in $\mathcal{S}(\mathbb{R}).$ And so is its Fourier transform $\hat{f}.$ Furthermore, $\hat{f}$ (as well as $f$), is of the form $p(\xi) e^{-\xi^2/4}$ for some polynomial $p(\xi)$ and is therefore so rapidly decreasing that $\xi \hat{f}(\xi) e^{-y\xi}$ is in $L^1(\mathbb{R})$ for all $y\in\mathbb{R}.$ It therefore satisfies both conditions so \eqref{analytic-fourier-inverse} applies.