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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 $f$?

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up vote 2 down vote accepted

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

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So the answer is that the integral is $T·\sum\limits_{n\geqslant0}\frac{f^{(n)}(0)}{(n!)^2}T^{n}$? – NikolajK Mar 13 '12 at 17:07
Caution: $\int_0^\infty f(x)\ e^{-x/T}\ dx$ might not converge for any $T$. – Robert Israel Mar 13 '12 at 17:27
Also, $\int_0^\infty x^n \ e^{-x/T}\ dx = n! T^{n+1}$, not $T^{n+1}/n!$. – Robert Israel Mar 13 '12 at 17:36
@RobertIsrael True. See revised version. – Did Mar 13 '12 at 17:43
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? – Aryabhata Mar 13 '12 at 18:38

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.

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