I have been reading David Angell's lovely book, *Irrationality and Transcendence in Number Theory*, which has given me some fresh insights even with some of the easier proofs. But the book reminds me of something that I've long been puzzled by, which the book hasn't cleared up for me. In the book's proof that $e^r$ is irrational for nonzero rational $r$, the starting point is to consider an integral of the form
$$\tag{\*}\label{star}\int\_0^r f(x) e^x \thinspace dx$$
for some suitably chosen polynomial $f(x)$. Lest you think this is an isolated trick used only in this one proof, let me mention that perhaps the shortest known proof of the irrationality of $\pi$ is based on the same idea (with $\sin x$ in place of $e^x$) as are even more general irrationality/transcendence proofs. My main question is:



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> Why would you think that multiplying by a polynomial and integrating is a fruitful idea here?
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To give a better idea of what I'm asking, let me make a few remarks about how far I've gotten with trying to answer my own question. It makes sense that for an irrationality proof, we should be looking for good approximations to $e^x$. Angell attributes the idea for his proof to Hermite, and it occurs to me that Hermite is also well known for his work on orthogonal polynomials. So maybe the thought process could go something like this: If Taylor series aren't working for us ([Fourier's proof of the irrationality of $e$](https://en.wikipedia.org/wiki/Proof_that_e_is_irrational#Fourier%27s_proof) does use the Taylor series, but if you try to mimic it naïvely to prove the irrationality of $e^r$, you run into some difficulties when $r>1$), then maybe we can try to approximate $e^x$ using orthogonal polynomials. This idea would at least get integration into the picture. So I have a side question:



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> Is there any historical evidence that Hermite's proof of the transcendence of $e$ was partially inspired by his work on orthogonal polynomials?
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One problem with this line of heuristic reasoning is that the best choices for the polynomial $f(x)$ seemingly have nothing to do with classical orthogonal polynomials. For the proof of the irrationality of $e^r$ or [Niven's proof of the irrationality of $\pi$](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-53/issue-6/A-simple-proof-that-pi-is-irrational/bams/1183510788.full), the appropriate choice of $f(x)$ turns out to be something of the form
$$\tag{\*\*}\label{starstar}f(x) = {x^n (a-bx)^n\over n!}$$
where $a/b$ is the rational number that we are trying to prove can't exist. Well, if you get as far as thinking that Formula \eqref{star} might be helpful, then it's not so hard, by staring at what you get via integration by parts, to "optimize" the choice of $f(x)$. So maybe that's all there is to it: motivated by a vague analogy with the theory of orthogonal polynomials, we are led to consider Formula \eqref{star}, and then we find the optimum choice of $f(x)$ for our purposes. However, I'm still left wondering if there's some general theory that can be used to justify that polynomials of the form \eqref{starstar} are interesting. Here's one observation to suggest that something more might be hiding behind the scenes. Suppose we evaluate
$$ \int\_0^1 {x^n(1-x)^n\over n!} e^x \thinspace dx.$$
We find that we get (some of) the continued fraction convergents for $e$—or more precisely, we get $p-qe$ for convergents $p/q$. (By contrast, if we truncate the Taylor series for $e$, we don't generally get convergents for $e$.) So a third question might be:



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> Is there a way to see a priori, without "just working it out," that we should expect the continued fraction convergents to emerge from $e$ in this manner?
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To prove irrationality, why integrate?
Here's an exposition of Niven's proof that makes the connection to orthogonal polynomials explicit. We start with an observation, easily proven by induction, that if $P\in \mathbb{Z}[x]$, then $\int\_0^\pi P(x)\sin x=Q(\pi)$, where $Q\in \mathbb{Z}[x]$, and $\deg Q\leq\deg P$. If we can find a sequence of polynomials $P\_n\in \mathbb{Z}[x]$, $\deg P\_n=n$, such that $\int\_0^\pi P\_n\sin x$ tends to zero super-exponentially, then we are done, since then the assumption that $\pi=a/b$ implies that $b^n\int\_0^\pi P\_n(x)\sin x$ is an integer (and non-zero for infinitely many $n$, since $\sin$ is not a polynomial).


Legendre polynomials on $(0,\pi)$ are perhaps the most natural try: $P\_n$ is orthogonal on $(0,\pi)$ to all polynomials of degree $\leq n-1$, and hence to the Taylor polynomial of $\sin x$ of degree $n-1$. The remainder decays super-exponentially, so that
$$
\left|\int\_0^\pi P\_n(x)\sin(x)\,dx\right|\leq\frac{\pi^{n+1}}{n!}||P\_n||\_2.
$$
From that point on, the only fact that you need to know about Legendre's polynomials to see that the proof is bound to work is that if you normalize them to have integer coefficients, the norm will be only exponentially large. Indeed, by Rodrigues' formula, we have for Legendre's polynomials on $(0,\pi)$
$$
P\_n(x)=\frac{\pi^{-n}}{n!}\frac{d^n}{dx^n}((x-\pi)x)^n=\frac{1}{n! a^n}\frac{d^n}{dx^n}((bx-a)x)^n,
$$
and those will have norm $\sqrt{\frac{\pi}{2n+1}}$ and have integer coefficients after multiplication by $a^n$, so we are done.


To obtain Niven's integral, plug in the above expression into $\int\_0^\pi P\_n(x)\sin (x)\, dx$ and integrate by parts $n$ times, replacing $\sin$ by $\cos$ if $n$ is odd. (Note that since all the derivatives of $(bx-a)^nx^n$ up to $(n-1)$-st vanish at the endpoints of the interval, there will be no boundary terms.) This gives an easy bound on the integral without reference to orthogonality and Taylor expansion, and also positivity. At the same time, it conceals that Legendre's polynomials were ever there :).


