1
$\begingroup$

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 $f$?

2 Answers 2

2

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 $\frac{T}{1-aT}$.

  • 0
    So the answer is that the integral is $T·\sum\limits_{n\geqslant0}\frac{f^{(n)}(0)}{(n!)^2}T^{n}$?2012-03-13
  • 0
    Caution: $\int_0^\infty f(x)\ e^{-x/T}\ dx$ might not converge for any $T$.2012-03-13
  • 0
    Also, $\int_0^\infty x^n \ e^{-x/T}\ dx = n! T^{n+1}$, not $T^{n+1}/n!$.2012-03-13
  • 0
    @RobertIsrael True. See revised version.2012-03-13
  • 0
    I am curious, I have seen this is many of your answers. You use $f:x\mapsto\mathrm e^{ax}$ instead of $f(x) = e^{ax}$. Is there any reason apart from like/dislike?2012-03-13
  • 0
    Indeed there is! The former denotes a *function* while the latter denotes a *number*. Here I want to define an object $f$ in the set $\mathbb R^{\mathbb R}$, not $f(x)$ in the set $\mathbb R$. Note that the complete notation would be $f:\mathbb R\to\mathbb R$, $x\mapsto\mathrm e^{ax}$, since the definition of a function includes a source set and a target set.2012-03-13
  • 0
    @DidierPiau: Ok I was hoping there might have been a less pedantic reason :-)2012-03-13
3

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.