Given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ we take the generalised Fourier transform $\hat{f}(w)=\int_{-\infty}^{+\infty}e^{iwx}f(x)dx$ where $w\in \mathbb{C}$.
Now assume, this transform will be defined only for some $\Im(w)\in B$, where $B$ is an interval on the real line, that is $\hat{f}(w)$ explodes for values of $\Im(w)$ outside $B$.
Since $\hat{f}(w)$ is defined in $A=\{w\in \mathbb{C}: \Im(w)\in B\}$, then we can define its transform on this set as $f(x)=\frac{1}{2\pi}\int_{i\beta-\infty}^{i\beta+\infty}e^{-iwx}\hat{f}(w)dw$. We can also say that on this set the function $f$ has an analytic complex extension, that is I can go back and forth between $f$ and $\hat{f}$ by the above formulae, in this set.
Problem
Instead of taking $\beta\in A$ in the inverse formula $f(x)=\frac{1}{2\pi}\int_{i\beta-\infty}^{i\beta+\infty}e^{-iwx}\hat{f}(w)dw$ I want to take $\beta $ outside of $A$. Now, because I am outside of the regularity strip given by $A$, I am confronted with the problem of singularities and so the inverse formula for $f(x)$ will have to include the residue afferent to the respective singularities. The singularities are the singularities of $\hat{f}(w)$. Now, the problem is that $\hat{f}(w)$ is not defined(??) for $\beta$ outside $A$. So, why am I even considering moving $\beta$ outside $A$?
Example Take a function $f(x)$ for which $\hat{f}(w)=\int_{-\infty}^{+\infty}e^{iwx}f(x)dx=\frac{C^{1+iw}}{w^2-iw}$ where the integral is defined only for $\Im{(w)}\in (1,\infty)$, for $\Im{(w)}$ outside of this interval the integrand $e^{iwx}f(x)$ explodes . Now, if I want to recover $f(x)$ by the inverse transform I can apply $f(x)=\frac{1}{2\pi}\int_{i\beta-\infty}^{i\beta+\infty}e^{-iwx}\hat{f}(w)dw$ with $\beta \in (1,\infty)$ and get $f(x)$.
The question is : can I take $\beta$ outside $(1,\infty)$ and get $f(x)$?
Well, $f(x)=\frac{1}{2\pi}\int_{i\beta-\infty}^{i\beta+\infty}e^{-iwx}\frac{C^{1+iw}}{w^2-iw}dw$ for $\beta \in (1,\infty)$, but for $\beta \in (0,\frac{1}{2})$, for example, we need to include the residue from the pole at $w=i$ of the integrand $e^{iwx}\frac{C^{1+iw}}{w^2-iw}$, that is $f(x)=-2\pi iRes(w=i)+\frac{1}{2\pi}\int_{i\beta-\infty}^{i\beta+\infty}e^{-iwx}\frac{C^{1+iw}}{w^2-iw}dw$,for $\beta \in (0,\frac{1}{2})$. But my problem is that $\hat{f}=\frac{C^{1+iw}}{w^2-iw}$ only for $\Im(w)\in (1,\infty)$, that is , is not defined for other values of $w$, like $\{w\in \mathbb{C}:\Im(w)\in(0,\frac{1}{2})\}$ as above.
Remark My problem is not related to why you have the extra term corresponding to the residue, but to why can I use $\frac{C^{1+iw}}{w^2-iw}$ instead of $\hat{f}$ outside of the region where $\hat{f}=\frac{C^{1+iw}}{w^2-iw}$, in the integral above.
I think I figure out the answer by myself. Just need some confirmation.
Basically, I was puzzled by the fact that I can move the contour into a region where $\hat{f}(w)$ was not defined. However, in the example, even though $\hat f$ is not equal to $\frac{C^{1+iw}}{w^2-iw}$ for $\beta$ outside $(1,\infty)$, this doesn't mean I can't integrate $\int^{i\beta+\infty}_{i\beta-\infty} e^{-iwx}\frac{C^{1+iw}}{w^2-iw}dw$ for $\beta$ outside $(1,\infty)$. The only thing is that the result will not be $f(x)$! Since the integral above for the contour that passes by $\beta$ outside $(1,\infty)$ will include a pole, I just have to detract the residue on that pole to get $f(x)$, that is $\int^{i\beta+\infty}_{i\beta-\infty} e^{-iwx}\frac{C^{1+iw}}{w^2-iw}dw$ for $\beta \in(1,\infty)$. Quite simple, just couldn't see it at first.
Could you please confirm the answer?