As I was sitting through a boring lecture rehashing basic techniques to solve ordinary differential equations, I began thinking about the Laplace transform and scribbled down a few ideas that I've copied below.

Consider the Laplace transform $\mathfrak{L}\{f\}:=\int_0^\infty e^{-st}f(t)~dt$ on a smooth function $f$ such that there exists some $s_0\in\mathbb{C}$ such that $e^{-s_0t}f(t)\to0$ as $t\to\infty$ (I believe it is sufficient to require $f$ is of exponential type). By integration by parts, it follows $$\mathfrak{L}\{f\}=s^{-1}f(0)+s^{-2}f'(0)+s^{-3}f''(0)+\dots=s^{-1}\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}$$

In other words, the Laplace transform behaves like a map from $f$ to the generating function of $(f^{(n)}(0))_{n\in\mathbb{N}}$ at $t=0$.

For example, consider a straightforward linear constant-coefficient ODE in $f$, say $f''-2f'+f=0$. 'Applying' the Laplace transform, we get: $$s^{-1}\sum_{n=0}^\infty f^{(n+2)}(0)~s^{-n}-2s^{-1}\sum_{n=1}^\infty f^{(n+1)}(0)~s^{-n}+s^{-1}\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}=0\\s^{-1}\left(s^2\sum_{n=2}^\infty f^{(n)}(0)~s^{-n}\right)-2s^{-1}\left(s\sum_{n=1}^\infty f^{(n)}(0)~s^{-n}\right)+s^{-1}\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}=0\\s\left(\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}-f(0)-f'(0)s^{-1}\right)-2\left(\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}-f(0)\right)+s^{-1}\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}=0\\(s^2-2s+1)s^{-1}\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}=sf(0)+f'(0)+2f(0)\\s^{-1}\sum_{n=0}^\infty f^{(n)}(0)~s^{-n}=\frac{sf(0)+f'(0)+2f(0)}{s^2-2s+1}$$

This so far agrees with the elementary result found using the basic properties of the Laplace transform and, in fact, with the standard method of solving recurrence relations using generating functions.

Indeed, it seems as though the Laplace transform is simply the method of generating functions applied to the sequence $(f^{(n)}(0))_{n\in\mathbb{N}}$ (over which the ODE reduces to a recurrence relation). However it seems almost necessary that for this interpretation to work, the function's behavior over the domain of interest must be fully described by its derivatives at $t=0$, i.e. $f$ must be analytic.

What can be concluded from the above? Is this a valuable approach to looking at the Laplace transform?

  • 2
    $\begingroup$ Using Laplace transforms to solve linear ODEs is analogous to using generating functions to solve linear recurrences. Here is a lecture from an MIT OCW course where the professor talks about this: youtube.com/watch?v=sZ2qulI6GEk $\endgroup$
    – Math1000
    Commented Jan 23, 2015 at 20:03
  • 3
    $\begingroup$ yes, that video is relatively standard and low-level, @Math1000; I've seen that video plenty of times over several years. It explains in a way how the Laplace and Z transforms (i.e. generating functions) are analogous in terms of basic properties, sure, but I've never seen a direct characterization of the method of Laplace transforms as itself the method of generating functions after identifying the Laplace transform with the map that takes analytic functions $f$ to a generating function in their derivatives $f(0), f'(0), f''(0)', \dots$ $\endgroup$
    – obataku
    Commented Jan 23, 2015 at 20:13

2 Answers 2


This is a special case of the moment generating property of the Laplace transform $^1$:

$$ \mathfrak{L}\Big\{f\Big\} = \int_{-\infty}^{\infty} \Big\{e^{-st} \Big\} f(t) \operatorname{dt} = \int_{-\infty}^{\infty} \Big\{\sum_{n=0}^{\infty}(-1)^n \frac{s^n t^n}{n!} \Big\} f(t) \operatorname{dt}$$ $$= \sum_{n=0}^{\infty}(-1)^n \frac{s^n}{n!} \int_{-\infty}^{\infty}t^n f(t)\operatorname{dt} = \sum_{n=0}^{\infty}(-1)^n \frac{s^n}{n!} m_n $$

where $m_n$ is the $n$-th moment of $f$ about $0$ (one assumes these exist).

In particular, if $f$ is the PDF of some random variable $X$, then $M_X (-s)$, the reflected moment generating function, is the Laplace transform of $f$.

see M.J. Dupré, «Transforms and Moment Generating Functions» for a more in-depth treatment. (PDF accessible as of 29/2/2015)

Note 1: The double-sided Laplace transform is used here because it supposedly translates better into the probabilistic case. It seems there is no significant difference between using the regular $\mathfrak{L}_+$ versus the two-sided $\mathfrak{L}$. However, I am not well-acquainted with its use in probability, so don't take my word for it.

Note: I just noticed that, expanding $f$ as a Taylor series, one arrives at the same equality:

$$ \mathfrak{L}\Big\{\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}t^n\Big\} = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\mathfrak{L}\Big\{t^n\Big\} = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\frac{n!}{s^{n+1}} = \frac{1}{s} \sum_{n=0}^{\infty} f^{(n)}(0)\frac{1}{s^n} $$

  • $\begingroup$ yeah, I'm familiar with moment generating functions and their relationship to the Laplace transform, but I don't see how the result follows from the fact $\mathfrak{L}\{f\}$ yields an alternating exponential generating function for the moments of $t$. The last note of yours is indeed one of the ways I thought of the result, though. $\endgroup$
    – obataku
    Commented Jan 29, 2015 at 17:55
  • $\begingroup$ moments of $f$ not $t$, sorry $\endgroup$
    – obataku
    Commented Jan 29, 2015 at 18:20

For what it's worth, I discovered shortly after posting this question that it's a rather well appreciated fact in the relevant fields that the Laplace transform indeed provides a map between ordinary- and exponential-type generating functions of a given sequence.

See also the technique of Borel summation.


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