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!

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 $$e^{-st}|g(t)|\le Me^te^{-st}\implies |e^{-st}g(t)|\le Me^{-t(s-1)}.$$
