We have the integral : $$\int_{1}^{\infty}\frac{\left \{ x \right \}-\frac{1}{2}}{x(\log x+z)}dx$$ $\left \{ x \right \}=$ fractional part of $x$, and $z$ is a complex variable whose real part is strictly positive. By the well-known summation formula : $$\sum_{a<n\leq b}f(n)=\int_{a}^{b}\left(f(x)+f^{'}(x)\left[\left \{ x \right \}-\frac{1}{2}\right]\right)dx+\frac{f(b)}{2}-\frac{f(a)}{2};\;\;\;\;\ a,b \in \mathbb{Z}^{+}$$ We have: $$\int_{1}^{\infty}\frac{\left \{ x \right \}-\frac{1}{2}}{x(\log x+z)}dx=\lim_{N\rightarrow\infty}\sum_{n=2}^{N}\log(\log(n)+z)+\frac{3}{2}\log(z)-e^{-z}\text{Ei}(z)+e^{-z}\text{Ei}\left(\log N+z\right)-\left(N+\frac{1}{2}\right)\log(\log N+z)$$ Where $\text{Ei}(\cdot)$ is the exponential integral function.
Now, we have: $$\sum_{n=2}^{N}\log(\log(n)+z)=\sum_{n=2}^{N}\log\log(n)+\sum_{n=2}^{N}\log\left(1+\frac{z}{\log n}\right)$$ Using similar arguments, we also have $$\lim_{N\rightarrow \infty}\sum_{n=2}^{N}\log\log(n)-\left(N+\frac{1}{2}\right)\log\log N +\text{li}(N)=k(=\text{constant})$$ Where $\text{li}(\cdot)$ is the logarithmic integral function.
Thus : $$\int_{1}^{\infty}\frac{\left \{ x \right \}-\frac{1}{2}}{x(\log x+z)}dx=k+\frac{3}{2}\log(z)-e^{-z}\text{Ei}(z)+\lim_{N\rightarrow \infty}\sum_{n=2}^{N}\log\left(1+\frac{z}{\log n}\right)+e^{-z}\text{Ei}\left(\log N+z\right)-\text{li}(N)+\left(N+\frac{1}{2}\right)\log\left(\frac{\log N}{z+\log N}\right)$$
My Idea to make the log-summation converge is to complement it with a sequence $a_{n}(z)$ such that :
$$\sum_{n=2}^{\infty}\log\left(1+\frac{z}{\log n}\right)-a_{n}(z)$$ is convergent as well as:
$$\lim_{N\rightarrow \infty}e^{-z}\text{Ei}\left(\log N+z\right)-\text{li}(N)+\left(N+\frac{1}{2}\right)\log\left(\frac{\log N}{z+\log N}\right)+\sum_{n=2}^{N}a_{n}(z)$$ What is the sequence $a_{n}(z)$ ?