Integral inequality in proof of irrationality of $\pi$

Question

I'm trying to follow along with my textbook's proof of $\pi$ being irrational. One step in the proof (which my book takes for granted) confuses me. It states that, for each positive integer $n$,

$$0 < \int\limits_0^\pi \frac{(px-qx^2)^n}{n!}\hspace{3pt}\sin(x) \hspace{3pt}dx < \pi \cdot \frac{p^n}{n!}$$

where $\pi = p/q$ (this is a proof by contradiction).

I was able to manipulate the integral, using simple algebra, as follows:

$$\frac{q^n}{n!}\int\limits_0^\pi x^n(\pi-x)^n\hspace{3pt}\sin(x) \hspace{3pt}dx$$

Any hints as to how I might prove this inequality from here?

Answer 1

On the interval $[0,\pi]$, we have $$ \left\lbrace \begin{align*} &0\le x(\pi-x)\le \frac{\pi^2}{4}\\[4pt] &0\le \sin(x) \le 1\\[4pt] \end{align*} \right. $$ hence \begin{align*} & \frac{q^n}{n!}\int_0^\pi x^n(\pi-x)^n\sin(x)\,dx \\[4pt] \le\; & \frac{q^n}{n!}\int_0^\pi \left(\frac{\pi^2}{4}\right)^n\,dx \\[4pt] =\; & \left(\frac{q^n}{n!}\right)\left(\frac{\pi^2}{4}\right)^n\left(\int_0^\pi 1\,dx\right) \\[4pt] =\; & \left(\frac{q^n}{n!}\right)\left(\frac{\pi^2}{4}\right)^n(\pi) \\[4pt] =\; & \left(\frac{(q\pi)^n}{n!}\right)\left(\frac{\pi}{4}\right)^n(\pi) \\[4pt] =\; & \left(\frac{p^n}{n!}\right)\left(\frac{\pi}{4}\right)^n(\pi) \\[4pt] < \; & \pi{\,\cdot\,}\frac{p^n}{n!} \\[4pt] \end{align*}

Answer 2

I think it should be $0 < \int\limits_0^\pi \frac{(px-qx^2)^n}{n!}\hspace{3pt}\sin(x) \hspace{3pt}dx <\frac{\pi^{n+1} p^n}{n!}$, (note the power of $\pi$) which follows because $$(px-qx^2)^n = x^n(p-qx)^n \le \pi^n p^n$$ Here, $x\le \pi$ implies $x^n\le \pi^n$ and also $0\le x\le p/q$ implies $0\le qx\le p$ and therefore $p-qx \le p$. My guess is backed up by this other question about this proof a simple proof that $\pi$ is irrational by Ivan Niven and the source paper [1].

See also Is there a proof that $\pi$ is an irrational number? for links to other proofs.

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^{[1] Niven, Ivan, A simple proof that $\pi$ is irrational, Bull. Am. Math. Soc. 53, 509 (1947). ZBL0037.31404.}

Answer 3

This is not an answer but it is too long for the comment section.

Being interested by $$I_n=\frac{1}{n!}\int\limits_0^\pi x^n(\pi-x)^n\hspace{3pt}\sin(x) \hspace{3pt}dx$$ I used the quadratic approximation given by @robjohn (have a look here) $$\sin(x) \sim \frac{31}{10\pi^2}(\pi-x)x+\frac{18}{5\pi^4}(\pi-x)^2x^2$$

This gives $$I_n=\frac{1}{80} (n+1) \pi ^{2 n+1} \left(\frac{49 \sqrt{\pi } }{4^n\,\Gamma \left(n+\frac{5}{2}\right)}-\frac{288 \Gamma (n+4)}{\Gamma (2 n+6)}\right)$$ which is quite accurate (relative error smaller than $0.1$% if $n>4$).

I do not think it could be useful for your problem but the approximation of the integral was nice compared to the unpleasant rigorous solution $$I_n=\left(\frac{\pi}{2 }\right)^{2 n+3} \Gamma (n+2) \,\,\, _2\tilde{F}_3\left(\frac{n+2}{2},\frac{n+3}{2};\frac{3}{2},n+\frac{3}{2},n+2; -\frac{\pi ^2}{4}\right)$$ where appears the regularized generalized hypergeometric function.