Let $f\in L_1(-\infty,\infty)$ be a Lebesgue-summable function on $\mathbb{R}$ and let $x\mapsto e^{\delta|x|}f(x)$ also be Lebesgue-summable on all the real line. I have added the condition that $f\in L_1(-\infty,\infty)$, but I think that if $e^{\delta|x|}f(x)$ is Lebesgue-summable (the only condition explicitly stated in the book, Kolmogorov-Fomin's, p. 430 here) on $\mathbb{R}$ then $f$ also is.
Kolmogorov-Fomin's Элементы теории функций и функционального анализа states (p. 430) that the complex function defined by $$g(z)=\int_{\mathbb{R}}f(x)e^{-ixz}d\mu_x$$where the integral is Lebesgue's, $\mu_x$ is Lebesgue linear measure and $z=\lambda+i\nu$, $\lambda,\nu\in\mathbb{R}$ is a complex variable, is analytic in $\{z\in\mathbb{C}:|\nu|<\delta\}$. I clearly see that the integral is the Fourier transform of $f(x)e^{\nu x}$.
By trying to prove the analiticity of $g$ I have convinced myself that, if $x\mapsto e^{\delta|x|}f(x)$ is Lebesgue-summable on $\mathbb{R}$, then $x\mapsto xf(x)$ also is. If that is correct, then I think we could use the Cauchy-Riemann equations and calculate the two partial derivatives with respect to $\lambda$ by using$^1$ the fact, where I call $F$ the Fourier transform, that $\frac{d}{d\lambda}F[e^{\nu x}f(x)](\lambda)=-iF[xe^{\nu x}f(x)]$. Nevertheless, I have no idea about how to calculate $\frac{\partial}{\partial\nu}\text{Re }g$ and $\frac{\partial}{\partial\nu}\text{Im }g$ (I do not even see why such derivatives exist).
Can anybody prove the analiticity of $g$ in $\{z\in\mathbb{C}:|\nu|<\delta\}$ either by using the Cauchy-Riemann equations or in some other way? I $\infty$-ly thank you!
$^1$To complete "what I have tried" I should say that I get the two following partial derivatives:$$\frac{\partial}{\partial\lambda}\text{Re }g(\lambda)=\int_{\mathbb{R}}xe^{\nu x}[\text{Im }f(x)\cos(x\lambda)-\text{Re }f(x)\sin(x\lambda)]d\mu_x$$ $$\frac{\partial}{\partial\lambda}\text{Im }g(\lambda)=\int_{\mathbb{R}}xe^{\nu x}[\text{Re }f(x)\cos(x\lambda)+\text{Im }f(x)\sin(x\lambda)]d\mu_x$$