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

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


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


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

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

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


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

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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.