Singularities on the boundary of the Laplace transform's region of convergence: poles or essential singularities?

Question

According to wikipedia, the Laplace transform $$F(s)=\int_0^{\infty} f(t)\exp(-s t) dt$$ converges in some region defined by $\Re(s)>s_0$, and converges to a holomorphic function in this region. Thus, (I think), $s_0$ must be the real part of the rightmost (largest real part) singularity of $F(s)$. My question is: is this singularity a pole, or could it be an essential singularity?

Answer 1

If $f\ge 0$ then yes there is a singularity on the real axis at the abscissa of convergence. It can be any kind of singularity (including essential singularity, branch point and natural boundary).

Otherwise no, sometimes $F(s)$ extends analytically to an entire function with the integral converging absolutely iff $\Re(s)>1$ and converging conditionally iff $\Re(s)>0$.

For example $\phi(t)=1_{t\in [0,1]}$ and $$f(t)=\sum_{n\ge 1} (-1)^{n+1} e^n \phi(e^n(t-\log n))$$

For $\Re(s) >0$ $$|(n+1)^{-s}|\le\int_0^\infty |e^n\phi(e^n(t-\log n)) e^{-st}|dt=\int_{\log n}^{\log n+e^{-n}} |e^{-st}|dt\le |n^{-s}|$$ and $$n^{-s}-\int_0^\infty e^n\phi(e^n(t-\log n)) e^{-st}dt$$ $$= \int_{\log n}^{\log n+e^{-n}} e^n(e^{-s\log n}-e^{-s\log t})dt= O(n^{-s} s e^{-n}) $$

Whence $\int_0^\infty |f(t)e^{-st}|dt\le \sum_{n\ge 1} |n^{-s}|$ converges iff $\Re(s)>1$ and $$F(s)-\sum_{n\ge 1} (-1)^{n+1} n^{-s}$$ extends to an entire function, and thus so does $F(s)$, and $$\lim_{T\to \infty} \int_0^T f(t)e^{-st}dt$$ converges iff $\Re(s)>0$.