Let $s=\sigma+it$ the complex variable where thus $i^2 =-1$, and $\sigma$ and $t$ are real numbers. Let $\mu(k)$ the Möbius function. It is possible determine the set of functions such that
$$M \left\{ f(t) \right\} =i\sum_{k=1}^\infty\frac{\mu(k)\log k}{k^{it}}\int_0^\infty\frac{f(t)}{k^s}dt$$ is defined?
I don't know if the rigth question is the previous since I believe that I need to provide also an abscissa $\sigma$. I thought of the previous transform when I've tried reproduce the shape of the Dirichlet series $$\frac{1}{\zeta(s)}=\sum_{k=1}^\infty\frac{\mu(k)}{k^s},$$ where is required that $\Re s=\sigma>1$. Then I can reproduce it so for $\sigma>1$ $$\sum_{k=1}^\infty i\frac{\log k}{k^{it}}\int_0^\infty e^{-it\log k}\frac{\mu(k)}{k^{\sigma}}1dt=\frac{1}{\zeta(s)},$$ that is: for $\sigma>1$ then $M \left\{ 1 \right\}=\frac{1}{\zeta(s)}$. I believe (I did computations for $M \left\{ 1 \right\}$, and also for $M \left\{ e^{-x} \right\}$) that my construction, when $g(s)=M \left\{ f(t) \right\}$ is defined, then $\frac{\partial}{\partial t}\Re g(s)=(\Im g(s))'$, and $\frac{\partial}{\partial t}\Im g(s)=-(\Re g(s))'$, where with the derivative $'$ we denote the derivative of arithmetic functions, that is multiply by $\log k$.
Then my question is
Question. It is possible determine the set of functions such that $$M \left\{ f(t) \right\} =i\sum_{k=1}^\infty\frac{\mu(k)\log k}{k^{it}}\int_0^\infty\frac{f(t)}{k^s}dt$$ is defined? It is neccesary require more conditions for which my problem is well-possed?
My computations were that it is obvious that $M \left\{ \cdot \right\} $ is linear and when the following is defined, one has $$ \left|M \left\{ f(t) \right\} \right| \leq \sum_{k=1}^\infty\frac{\log k}{k^\sigma} \int_0^\infty \left| f(t) \right|dt. $$
Seem that one needs conditions for the definition of such transform since $\int_0^\infty \left| f(t) \right|dt$ could be not defined.
Thus I don't know if my previous questions makes sense, and I would like to know if it is possible define the previous transform, as I've said I've computed also the transform of a negative expoenential. Thanks.