Let $f(x)$ be a real-valued function defined in $(0, \infty)$. I am curious what kind of $f(x)$ has the following representations:
$$
f(x) = \sum\_{j=0}^\infty a\_j e^{-jx}, \quad \forall x \in (0, \infty).
$$


My initial thought was that consider the transformation $s = e^{-x}, s \in (0, 1)$, then we want to find $\{a\_j\}\_{j=1}^\infty$ such that
$$
f(-\log s) = \sum\_{j=0}^\infty a\_j s^j, \quad \forall s \in (0, 1)
$$


That means we may consider the Taylor expansion for $f(-\log s)$ at $s=0$, but then $-\log s = \infty$, for which the Taylor series is not well defined (at infinity).


So is there a method to determine $a\_j$ if $f(x)$ can be represented as an exponential sum?



What kind of functions can be represented as infinite linear combinations of exponential functions?
$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$You were almost there.


Indeed, suppose that
$$f(x)=\sum\_{j=0}^\infty a\_j e^{-jx}\in\R \quad \forall x\in(0,\infty). \tag{1}\label{1}$$
Then
$$g\_f(s):=f(-\ln s)=\sum\_{j=0}^\infty a\_j s^j\in\R \quad \forall s\in(0,1).$$
So, the radius of convergence of the power series $\sum\_{j=0}^\infty a\_j z^j$ is $1$. So, $g\_f$ can be extended to the analytic function $G\_f$ on the open unit disc
$D$ in $\C$ given by the formula $G\_f(z)=\sum\_{j=0}^\infty a\_j z^j$ for $z\in D$.


Vice versa, if the function $g\_f$ can be extended to an analytic function $G\_f$ on
$D$, then $G\_f(z)=\sum\_{j=0}^\infty a\_j z^j$ with $a\_j=G\_f^{(j)}(0)/j!$ and hence \eqref{1} holds.


Thus, \eqref{1} holds for some $a\_j$'s if and only if the function $g\_f$ can be extended to an analytic function on $D$.


Similarly, \eqref{1} holds for some $a\_j$'s if and only if the function $g\_f$ can be extended to a real-analytic function on the interval $(-1,1)$.




---


For an illustration, suppose that $f(x)=\dfrac1{1-e^{-x}}$ for $x\in(0,\infty)$. Then $g\_f(s)=f(-\ln s)=\dfrac1{1-s}$ for $s\in(0,1)$. Clearly, the function $g\_f$ can be extended to the analytic function $G\_f$ on $D$ given by the formula
$$G\_f(z):=\dfrac1{1-z}=\sum\_{j=0}^\infty a\_j z^j=\sum\_{j=0}^\infty z^j$$
with $a\_j=G\_f^{(j)}(0)/j!=1$. So,
$$f(x)=\sum\_{j=0}^\infty a\_j e^{-jx}=\sum\_{j=0}^\infty e^{-jx}$$
for $x\in(0,\infty)$.


