I'm going to be teaching a Master's level analysis course(measure theory, Lebesgue integration, Banach and Hilbert spaces, and if there's time, some spectral or PDE stuff) in the fall. My problem is that while roughly half of the students will actually be using analysis in their further work, the rest of them are going to specialize in combinatorics, and while I want to convince them that they should know this stuff as part of their general mathematical culture, I'd also like to try to connect it to what they'll be working on. So far, I've found survey articles on applications of Ramsey theory to Banach spaces, and applications of harmonic analysis to additive number theory, but I was wondering whether anyone had some suggestions of references for applications of classical analysis to old-fashioned, classical combinatorics. (I realise that this is a pretty tall order, as on many levels, these two fields are at antipodes.)
PS: I'm planning on talking about probability measures on discrete spaces, but I don't think that will convince the combinatorics people that hacking through the construction of the Lebesgue integral could have a practical payoff someday for them.
Best Answer
Here's a nice application of measure theory, precisely, of the the theory of orthogonal polynomials, to a classic problem of counting derangements.
Problem: How many anagrams with no fixed letters of a given word are there?
For instance, for a word made of only two different letters, say $n$ letters $A$ and $m$ letters $B$, the answer is, of course, 1 or 0 according whether $n = m$ or not, for the only way to form an anagram without fixed letters is, exchanging all the $A$ with $B$, and this is possible if and only if $n=m$.
In the general case, for a word with $n_1$ letters $X_1$, $n_2$ letters $X_2$, ..., $n_r$ letters $X_r$, you will find (after the proper use of the inclusion-exclusion formula) that the answer has the form of a sum of products, that looks very much like the expansion of a product of sums, yet it is not. It is not, exactly because of the presence of terms $k!$, that would formally make a true expansion of a product of sums, if only they where replaced by corresponding terms $x^k$. This suggests to express them with the Eulerian integral $k!=\int_0^\infty x^ke^{-x}dx$, with the effect that the said expression becomes an integral (with the weight $e^{-x}$) of a true product of sums: precisely,
$$\int_0^\infty P_{n_1} (x) P_{n_2}(x)\cdots P_{n_r}(x)\, e^{-x}\, dx,$$
with a certain sequence of polynomials $P_n$, where $P_n$, has degree $n$. But the above answer for the case $r=2$ gives an orthogonality relation, whence the $P_n$, are the Laguerre polynomials, (up to a sign that is easily decided). Note that in the case with no repeated letters, all $n_i=1$, one finds again the more popular enumeration of permutations without fixed points.
Disclaimer: I partially copied this from wikipedia; it's me who wrote it there. The above is my personal amateur's solution, and possibly differs slightly from the vulgata. An on-line reference, with generalizations of the problem, is e.g.
Weighted derangements and Laguerre polynomials, D.Foata and D.Zeilberger, SIAM J. Discrete Math. 1 (1988) 425-433.