Explanation for $\mathcal{L} (\theta(e^t))$ is holomorphic on $Re(s) > 1$

Question

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!

Answer 1

The Laplace transform is $$\mathcal L(\vartheta(e^t))=\int_0^\infty e^{-st}\mathcal O(e^t)\,dt=\int_0^\infty\mathcal O(e^{-t(s-1)})\,dt$$ which converges whenever $\Re(s)>1$.

The last equality is justified by defining $$g(t)=\mathcal O(e^t)\iff|g(t)|\le Me^t\quad\forall t\ge t_0$$ and $M,t_0\in\Bbb R$, so $$e^{-st}|g(t)|\le Me^te^{-st}\implies |e^{-st}g(t)|\le Me^{-t(s-1)}.$$

Answer 2

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.