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en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Cartwright.27s_proof
Harold Jeffreys wrote that this proof was set as an example in an exam at Cambridge University in 1945 by Mary Cartwright, but that she had not traced its origin. It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University.
Consider the integrals
$$I_{n}(x)=\int _{{-1}}^{1}(1-z^{2})^{n}\cos(xz)\,dz,$$
where $n$ is a non-negative integer.
Two integrations by parts give the recurrence relation
$$x^{2}I_{n}(x)=2n(2n-1)I_{n-1}(x)-4n(n-1)I_{n-2}(x).\qquad (n\geq 2)$$
If
$$J_{n}(x)=x^{2n+1}I_{n}(x),$$
then this becomes
$$J_{n}(x)=2n(2n-1)J_{{n-1}}(x)-4n(n-1)x^{2}J_{{n-2}}(x).$$
Furthermore, $J_0(x) = 2\sin(x)$ and $J_1(x) = −4x \cos(x) + 4\sin(x)$. Hence for all $n ∈ \Bbb Z^+$,
$$J_{n}(x)=x^{2n+1}I_{n}(x)=n!{\bigl (}P_{n}(x)\sin(x)+Q_{n}(x)\cos(x){\bigr )},$$
where $P_n(x)$ and $Q_n(x)$ are polynomials of degree $≤ n$, and with integer coefficients (depending on $n$).
Take $x = π/2$, and suppose if possible that $π/2 = a/b$, where $a$ and $b$ are natural numbers (i.e., assume that $π$ is rational). Then
$$ \frac{a^{2n+1}}{n!}I_n\left(\frac\pi2\right) = P_n\left(\frac\pi2\right)b^{2n+1}. $$
The right side is an integer. But $0 < I_n(π/2) < 2$ since the interval [−1, 1] has length 2 and the function that is being integrated takes only values between 0 and 1. On the other hand,
$${\frac {a^{2n+1}}{n!}}\to 0\quad {\text{ as }}n\to \infty .$$
Hence, for sufficiently large $n$
$$ 0 < \frac{a^{2n+1}I_n\left(\frac\pi2\right)}{n!} < 1, $$
that is, we could find an integer between 0 and 1. That is the contradiction that follows from the assumption that $π$ is rational.
This proof is similar to Hermite's proof. Indeed,
\begin{aligned}J_{n}(x)&=x^{2n+1}\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\[5pt]&=2x^{2n+1}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\[5pt]&=2^{n+1}n!A_{n}(x).\end{aligned}
However, it is clearly simpler. This is achieved by omitting the inductive definition of the functions $A_n$ and taking as a starting point their expression as an integral.
反设 $\pi\in\mathbb{Q}$ ,那么将存在两个正整数 $u,v\in\mathbb{Z^+}$ ,使得 $\pi=\frac{u}{v}$ 。接下来我们将会构造一系列的积分，至于为什么这么构造，往后看。
\begin{aligned}
I_0&=2\\
I_1&=4v^2\\
&\vdots\\
I_n&=\frac{v^{2n}}{n!}\int_{0}^{\pi}x^n(\pi-x)^n\sin(x)\mathrm{d}x
\end{aligned}
这个积分看起来天外飞仙一样，但是注意到如果我们用分部积分来展开通项，就可以观察到这个性质
\begin{aligned}I_n=&\frac{v^{2n}}{n!}\int_{0}^{\pi}x^n(\pi-x)^n\sin x\mathrm{d}x
\\ =&\frac{v^{2n}}{n!}\int_{0}^{\pi}x^n(\pi-x)^n \mathrm{d}(-\cos x)
\\ =&\frac{v^{2n}}{n!}\left(-x^n(\pi-x)^n\cos x\big|_{x=0}^{\pi}+\int_{0}^{\pi}(x^n(\pi-x)^n)'\cos x\mathrm{d}x\right)\end{aligned}
注意到前面一项在0和$π$是都是0，所以只剩下分部积分的第二项，求导之后自然分裂成两部分，提取公因式得到
$$\frac{v^{2n}}{n!}\left(\int_{0}^{\pi}nx^{n-1}(\pi-x)^{n-1}(\pi-2x)\cos x\mathrm{d}x\right)$$
然后我们就可以整理一下，连同 $\pi^2=\frac{u^2}{v^2}$ 代入，我们得到（终于得到）一个重要的递推式
\begin{aligned}
I_n&=2(2n-1)v^2\frac{v^{2(n-1)}}{(n-1)!}\int_{0}^{\pi}x^{n-1}(\pi-x)^{n-1}\sin x\mathrm{d}x-u^2v^2\frac{v^{2(n-2)}}{(n-2)!}\int_{0}^{\pi}x^{n-2}(\pi-x)^{n-2}\sin x\mathrm{d}x
\\&=2(2n-1)v^2I_{n-1}-u^2v^2I_{n-2}
\end{aligned}
也就是说，这一系列的积分有简单的递推公式（这是初等数学），由于前两项0和 $4v^2$ 都是整数，递推的这些系数根据假设也是整数，所以我们得到这一系列的积分结果都应该是整数。
好吧，整数就整数吧，有什么问题吗？
有问题，因为我们可以对积分做一个简单的放缩
\begin{aligned}
I_n&=\frac{v^{2n}}{n!}\int_{0}^{\pi}x^n(\pi-x)^n\sin(x)\mathrm{d}x
\\& \leq\frac{v^{2n}}{n!}\int_{0}^{\pi}\pi^n\pi^n\cdot 1\mathrm{d}x
\\ &=\frac{v^{2n}\pi^{2n+1}}{n!}=\pi\cdot\frac{u^{2n}}{n!}
\end{aligned}
分子是一个指数函数，分母是一个阶乘（也就是伽马函数），我们有
$$\lim_{n\to\infty}\frac{u^{2n}}{n!}=0$$
也就是说，一个都是正整数的数列，却必须极限是0（好了，什么是极限？初等数学？）这是不可能的，所以不存在这样的正整数u和v，证明完毕。
（这个证明不是我想的，是Cartwright的，在1945年还出给剑桥学生做期末考试题，但是其证明的构造技巧非常精妙， 也是我个人最喜欢的一个，其余还有用连分数等证明方法的，但是都脱离不开无限数列或者级数乃至积分的概念。） |
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