Is there a convenient formula for this suppressed integral? Given that 
$$\int_0^\infty x^n\  \text{e}^{-\frac{x}{T}}\text{d}x\ \propto\  T^{n+1},$$
with n integer, is there are formula in terms of taylor expandable functions $f$ for 
$$\int_0^\infty f(x)\ \text{e}^{-\frac{x}{T}}\text{d}x?$$
That is, can I give a formula for this integral which only contains algebraic substitutions and/or maybe derivatives of $f$?
 A: Assuming $f(x)$ is exponentially bounded on $(0,\infty)$, say $|f(x)| \le C e^{B x}$ for some real constants $B$ and $C$, then
$\int_0^\infty f(x) e^{-x/T}\ dx = F(1/T)$ where $F$ is the Laplace transform of $f$.  It is defined and analytic in the region $\text{Re}(1/T) > B$.  However, it may or may not extend to an analytic function in a neighbourhood of $T=0$.  Watson's lemma says that if 
$f(x) = \sum_{n=0}^\infty a_n x^n$, $F(1/T) \sim \sum_{n=0}^\infty a_n \ n!\ T^{n+1}$ in the sense of an asymptotic series as $T \to 0$ in a sector of the right half plane.
A: If $f(x)=\sum\limits_{n\geqslant0}a_nx^n$ converges for every $x\geqslant0$, then 
 $a_n=\frac1{n!}f^{(n)}(0)$ for every $n\geqslant0$. Furthermore,
$$
\int_0^{+\infty}f(x)\mathrm e^{-x/T}\mathrm dx=\sum\limits_{n=0}^{+\infty}a_n\int_0^{+\infty}x^n\mathrm e^{-x/T}\mathrm dx=\sum\limits_{n=0}^{+\infty}n!\,a_nT^{n+1}=\sum\limits_{n=0}^{+\infty}f^{(n)}(0)T^{n+1},
$$
if the series in the RHS is absolutely convergent. 
Example: if $f:x\mapsto\mathrm e^{ax}$, then $f^{(n)}(0)=a^n$ for every $n\geqslant0$, the integral in the LHS converges if and only if $aT\lt1$, and then, the RHS is $\frac{T}{1-aT}$.
