Explanation for $\mathcal{L} (\theta(e^t))$ is holomorphic on $Re(s) > 1$ Let, $\theta(x)=\sum_{p\leq x} \log p$, it is known, $\theta(x) \sim x$, therefore, $\theta(e^t) = O(e^t)$. Recall the definition of the Laplace transform.  I found in a lecture note (click and go to Page 6) that  the Laplace transform$\mathcal{L} (\theta(e^t))$ is holomorphic on $Re(s) > 1$. See the below picture- 

Can someone please explain how $\mathcal{L} (\theta(e^t))$ is inferred holomorphic on $Re(s) > 1$ automatically, in detail?
Thanks!

There are 3 ways

*

*$\sum_{p \le e^t} \log p \le  te^t$ thus $$F_n(s)=\int_0^n (\sum_{p \le e^t} \log p)e^{-st}dt$$ converges uniformly to $F(s)$ on $\Re(s) \ge 1+\epsilon$. Because each $F_n$ is analytic it implies the limit is analytic.


*Both $$F(s) = \int_0^\infty (\sum_{p \le e^t} \log p)e^{-st}dt,F'(s) = -\int_0^\infty (\sum_{p \le e^t} \log p)te^{-st}dt$$
converge on $\Re(s) > 1$ thus $F(s)$ is complex differentiable  (holomorphic) on $\Re(s) > 1$. From the Cauchy integral formula it implies it is analytic.


*For $|s-a|< \Re(a)-1$
$$F(s) = \sum_{k=0}^\infty \frac{(a-s)^k }{k!}\int_0^\infty (\sum_{p \le e^t} \log p)t^k e^{-at}dt$$ converges absolutely thus it is analytic there
Similar ideas apply when investigating the abscissa of convergence of Dirichlet series.
