Suppose that you have an exponential generating function.: $E(z)=\sum_{n=0}^{\infty} \frac{a_{n}z^{n}}{n!}$, and that the definition of $a_{n}$ can be reasonably extended to noninteger arguments. (the Catalan numbers $C_{n}$, would be written in terms of the Gamma function thusly: $C_{n} = \frac{\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+1)}$, for instance), what then is the combinatorial significance of this integral:

$$U(z)=\int_{0}^{\infty} \frac{a_{v}z^{v}dv}{\Gamma(v+1)}$$ ?

  • $\begingroup$ You maybe should write $a(v)$; the question is then, how to get a nice form for $a$ if you have $U$? $\endgroup$
    – Raphael
    Nov 11, 2010 at 16:28
  • 5
    $\begingroup$ Is there some reason to expect that this integral has combinatorial significance? $\endgroup$ Nov 11, 2010 at 18:20
  • $\begingroup$ What is the motivation for such an integral? $\endgroup$
    – Mitch
    Mar 9, 2011 at 22:26
  • $\begingroup$ Do you have one example where you can describe $U$ explicitely? $\endgroup$ Mar 10, 2011 at 0:18
  • $\begingroup$ I looked at both the definition of the gamma function $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\, dt$ and the Fransén–Robinson constant $F = \int_{0}^\infty \frac{1}{\Gamma(x)}\, dx$, and thought that the way that $e^{-t}$ is in the Gamma function itself sort of resembles the way that the gamma function is used in the constant. This got me thinking about the parallels between summation and integration. And I know that there are many sequences with easy interpolations to the reals. So I wondered what would happen if one, instead of hanging the sequence on a generating function. (cont). $\endgroup$ Mar 10, 2011 at 18:31

1 Answer 1


'Combinatorial' usually refers to finite or discrete objects so using an integral really is going quite the other direction to an analytic one. That is, a combinatorial interpretation of an integral, as is, will most likely just be forced or convoluted (I'm sure it is possible under some intellectual circumstances, it's just not obvious how).

So maybe one would want just an unqualified interpretation of what it means when one takes a gf (of an already combinatorial situation) and replaces the summation symbol with an integral to get an analytic situation (which accords more with an analytic continuation or interpretation).

Then, the question really might be 'What is the analytic significance of the integral?' or possibly 'What is an analytic analog of a generating function? If a gf corresponds to a particular interpretation, what can we say, interpretively, about the symbolic transform from a summation to an integral?'

For the above example, this just isn't obvious, and the integral you give seems to be unknown (even if $a_n = 1$). so I'll try to say something in general about gfs that might help.

A gf is a way to capture a function on the naturals 'by other means'; the natural number exponents of the gf mark the values of the original function. So by trying to use an integral, you're trying to interpret the exponent as a real (rather than an integer). The furthest I've seen gfs generalized is to Puiseux series which allow rational exponents (but with some restrictions akin to a gf, some kind of arithmetical progression in the exponent).

The only appropriate 'conversion' I can think of is something like a Fourier or Laplace transform; those will convert a function to another 'domain' where manipulation can occur, and then convert back to the functional domain.

Edited out (the following is not what the OP is looking for):

The coefficient of $x^n/n!$ in an egf is usually combinatorially interpreted as the number of labeled sets of size $n$ (with a given structure).

A more likely integral related to an egf is to integrate (formally, forget the constant) the entire egf with respect to the egf variable, which has an immediate calculation:

$$ \int E(z) \ dz = \sum_{n=0}^{\infty} \int\frac{a_{n}z^{n}}{n!}\ dz = \sum_{n=1}^{\infty} \frac{a_{n}z^{n+1}}{(n+1)!} = \sum_{n=1}^{\infty} \frac{a_{n-1}z^n}{n!}$$.

All this does is, in a sense, shift the function: $\int E(z)$ is the egf of $a_{n-1}$.

Of course, the derivative of the egf shifts the other direction.

For ordinary generating functions (ogf), the derivative usually means combinatorially that you are 'pointing to' (distinguishing, choosing) one particular element out of a structure of size $n$ (and the integral 'undoes' the pointing). For egfs, the labeling provides a sort of 'pointing' already for every object, so the $n$ multiplier is irrelevant.

I realize I have answered a different question than you have asked, in the hopes that it is what you are really after. You are asking, given an egf for a function over naturals, to presume the existence of an analytic continuation of the original function, and then integrate over that analytic continuation (along with $z^n/n!$ also interpreted over reals). For the moment, I really can't say anything there other than it looks like some kind of fractional derivative.

  • $\begingroup$ This is not what I'm after. $\endgroup$ Mar 10, 2011 at 18:36

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