Let $$F(x)=\frac{ \left\{ x \right\} }{e^{\sqrt{x}}},$$ be supported on $ \left( 0,\infty \right) $, where $ \left\{ x \right\} $ is the fractional part function. Then $F\in L^2(0,\infty)$ and the Paley-Wiener theorem proves that $$f(s)=\int_0^\infty \left\{ x \right\} e^{ixs-\sqrt{x}}dx \\$$ is an holomorphic function for $\Im s>0$. As said the Wikipedia's article on the other hand Plancherel's theorem states that $$\int_{-\infty}^{\infty} \left| f(\xi+i\eta) \right|^2d\xi\leq \int_0^\infty \frac{ \left\{ x \right\}^2 }{e^{2\sqrt{x}}}dx, $$ and by the dominated convergence one has $$\lim_{\eta\to0^{+}}\int_{-\infty}^{\infty} \left| f(\xi+i\eta)-f(\xi) \right|^2d\xi =0.$$
Question. Was right the application of Paley_Wiener, this is does satisfy $F(x)$ the hypothesis of the theorem? Is there some relationship between the deduction from Plancherel and the mean value formula for Dirichlet series? Can you provide us some hints or computations to show that truly using dominated convergence we can get the third claim for my example? Thanks in advance.
To clarify the second doubt, I say that the shape of the integral is $\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T$ and the integration is over $\eta$, thus I thought that perhaps with a suitable change of variables and doing complex integration perhaps it is possible to get a series likes Apostol's Theorem 12.16, in Apostol, Introduction to Analytic Number Theory (Springer). For the third doubt, it is only to see more good computations, I say that if I have the definition of dominated convergence in a hand, what are the key computations or reasonings to see that the deduction is feasible for the holomorphic Fourier transform $f$.
The Parseval identity (a.k.a. Plancherel's Theorem) for the Fourier transform is $$ \int_{-\infty}^{\infty}|f(t)|^2dt = \int_{-\infty}^{\infty}|\hat{f}(t)|^2 dt,\;\;\; f \in L^2(\mathbb{R}). $$ I assume you're familiar with this theorem. If you have a function $f\in L^2(\mathbb{R})$ that is supported in $[0,\infty)$ only, then the Fourier transform extends into the Complex plane \begin{align} \hat{f}(u+iv) = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}f(t)e^{-ixu-ix(iv)} = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}f(t)e^{-ixu}e^{xv}dx, \;\;\; v < 0. \end{align} (Of course you can use the inverse Fourier transform instead of the forward transform of $f$ and extend into the complex plane where $v > 0$.) You can also view this as $$ \hat{f}(u+iv)=\widehat{fe^{xv}}(u),\;\;\; v < 0. $$ Then you can apply the Plancherel Theorem: \begin{align} \int_{-\infty}^{\infty}|\hat{f}(u+iv)-\hat{f}(u)|^2du & = \int_{-\infty}^{\infty}|\widehat{fe^{xv}}(u)-\hat{f}(u)|^2du \\ & = \int_{-\infty}^{\infty}|\widehat{fe^{xv}-f}|^2du \\ & = \int_{0}^{\infty}|f(x)e^{xv}-f(x)|^2dx \\ & = \int_{0}^{\infty}|f(x)|^2(e^{xv}-1)^2dx,\;\;\; v < 0. \end{align} Now you can see that the dominated convergence theorem can be used to prove that the far right side tends to $0$ as $v\rightarrow 0$ through negative values of $v$. For the other part of what you were looking at: \begin{align} \int_{-\infty}^{\infty}|\hat{f}(u+iv)|^2du & = \int_{-\infty}^{\infty}|\widehat{fe^{xv}}(u)|^2du \\ & = \int_{-\infty}^{\infty}|f(x)e^{xv}|^2dx \\ & \le \int_{-\infty}^{\infty}|f(x)|^2dx,\;\;\; v < 0. \end{align} The only requirement for $f$ is that it is $0$ for $t < 0$ and is in $L^2[0,\infty)$. Your function definitely satisfy those conditions because $e^{-\sqrt{x}}$ is square integrable, and your function is bounded by this function. To see that the $L^2$ norm is finite, let $r=\sqrt{x}$, $x=r^2$, $dx=2rdr$ in the following: \begin{align} \int_{0}^{\infty}(e^{-\sqrt{x}})^2dx & = \int_{0}^{\infty}e^{-2\sqrt{x}}dx = \int_{0}^{\infty}2e^{-2r}rdr < \infty. \end{align}