# Holomorphic extension of the Laplace transform

Let $u \in L^1_{\mbox{loc}}((0,+\infty))$ be such that $e^{-\lambda t} u(t) \in L^1((0,+\infty))$ for some $\lambda >0$. Let $\mathcal{L}[u]$ be the Laplace transform of $u$, that is $$\mathcal{L}[u](s) = \int_0^{+\infty}e^{-st}u(t)\,dt \qquad\mbox{ for all}\, s \in \mathbb{C} \,\mbox{ s.t. } \mbox{Re}(s)>\lambda.$$

If $\mathcal{L}[u]$ is holomorphic in $\mbox{Re}(s)>\lambda$ and admits a holomorphic extension $F(s)$ to the semi-plane $\mbox{Re}(s)>\lambda_1$ with $\lambda_1 < \lambda$, then is it true that $e^{-\lambda_2 t} u(t) \in L^1((0,+\infty))$ for all $\lambda_2 > \lambda_1$ and $\mathcal{L}[u](s) = F(s)$ in $\mbox{Re}(s)>\lambda_1$? Do you have a proof or a reference for this fact?

This is not the case, there are functions whose Laplace transform is entire (or rather, whose Laplace transform is the restriction of an entire function to a half-plane) but the integral converges only for $\operatorname{Re} s > \lambda_0$. I will give an example constructed from the Dirichlet $\eta$ function. Using

$$\frac{1}{n^s} = s\int_n^{+\infty} \frac{dx}{x^{s+1}} = s\int_1^{+\infty} \frac{\chi_{[n,+\infty)}(x)}{x^{s+1}}\,dx$$

(where $n^s$ and $x^{s+1}$ are formed using the real logarithm of $n$ and $x$) one obtains an integral representation for a Dirichlet series

$$F(s) = \sum_{n = 1}^{+\infty} \frac{a_n}{n^s}\,, \tag{1}$$

namely

$$F(s) = s\int_1^{+\infty} \frac{A(x)}{x^{s+1}}\,dx \tag{2}$$

where

$$A(x) = \sum_{n \leqslant x} a_n\,.$$

In the half-plane of absolute convergence of $(1)$ we can interchange summation and integration in

$$F(s) = \sum_{n = 1}^{+\infty} s\int_1^{+\infty} \frac{a_n \chi_{[n,+\infty)}(x)}{x^{s+1}}\,dx$$

by the dominated convergence theorem, and

$$\sum_{n = 1}^{+\infty} a_n\chi_{[n,+\infty)}(x) = A(x)\,.$$

By the identity theorem, $(2)$ holds on every half-plane where both sides are defined. Now consider the entire function

$$F(s) = \eta(s) - \eta(0)\,.$$

For $\operatorname{Re} s > 0$ this is given by the Dirichlet series

$$\frac{1}{2} + \sum_{n = 2}^{+\infty} \frac{(-1)^{n-1}}{n^s}$$

and consequently by the integral

$$s\int_1^{+\infty} \frac{B(x)}{x^{s+1}}\,dx$$

where

$$B(x) = \frac{1}{2}\cdot (-1)^{\lfloor x\rfloor - 1}$$

for $x \geqslant 1$. Making the substitution $x = e^t$ we find that for $u(t) = B(e^t)$ we have

$$\mathcal{L}[u](s) = \int_0^{+\infty} u(t)e^{-st}\,dt = \int_1^{+\infty} \frac{B(x)}{x^{s+1}}\,dx = \frac{\eta(s) - \eta(0)}{s}$$

for $\operatorname{Re} s > 0$. So the Laplace transform of $u$ has a holomorphic continuation to the entire plane, but since $\lvert u(t)\rvert = \frac{1}{2}$ for all $t \geqslant 0$ we have $u(t)e^{-\lambda t} \in L^1((0,+\infty))$ if and only if $\lambda > 0$. And the integral also doesn't exist as an improper Riemann integral for $\operatorname{Re} s \leqslant 0$.

However, things are different if $u$ is eventually nonnegative (or nonpositive). If $u \in L^1_{\text{loc}}((0,+\infty))$ and there is a $t_0$ such that $u(t) \geqslant 0$ for $t \geqslant t_0$, then

$$\lambda_0 = \inf \: \Biggl\{ \lambda \in \mathbb{R} : \int_{t_0}^{+\infty} u(t)e^{-\lambda t}\,dt < +\infty\Biggr\}$$

is a singularity (not necessarily a pole) of $\mathcal{L}[u]$. (This is strongly related to Landau's lemma for Dirichlet series with nonnegative coefficients.)