Ivan Niven published this proof in 1947. He used the characterization of \pi as the smallest positive
zero of the
sine function. Suppose that \pi is rational, i.e. \pi = a/b for some integers a and b which may be taken
without loss of generality to both be positive. Given any positive integer n, we define the polynomial function: : f(x) = \frac{x^n(a - bx)^n}{n!} and, for each x \in \R let :F(x) = f(x)-f''(x)+f^{(4)}(x)-\cdots+(-1)^n f^{(2n)}(x).
Claim 1: F(0) + F(\pi) is an integer.
Proof: Expanding f as a sum of monomials, the coefficient of x^k is a number of the form c_k /n! where c_k is an integer, which is 0 if k Therefore, f^{(k)}(0) is 0 when k and it is equal to (k! / n!) c_k if in each case, f^{(k)}(0) is an integer and therefore F(0) is an integer. On the other hand, f(\pi-x) = f(x) and so (-1)^kf^{(k)}(\pi-x) = f^{(k)}(x) for each non-negative integer k. In particular, (-1)^kf^{(k)}(\pi) = f^{(k)}(0). Therefore, f^{(k)}(\pi) is also an integer and so F(\pi) is an integer (in fact, it is easy to see that Since F(0) and F(\pi) are integers, so is their sum.
Claim 2: : \int_0^\pi f(x)\sin(x)\,dx=F(0)+F(\pi)
Proof: Since f^{(2n + 2)} is the zero polynomial, we have : F'' + F = f. The
derivatives of the
sine and
cosine function are given by sin' = cos and cos' = −sin. Hence the
product rule implies : (F'\cdot\sin{} - F\cdot\cos{})' = f\cdot\sin By the
fundamental theorem of calculus : \left. \int_0^\pi f(x)\sin(x)\,dx= \bigl(F'(x)\sin x - F(x)\cos x\bigr) \right|_0^\pi. Since \sin 0 = \sin \pi = 0 and \cos 0 = - \cos \pi = 1 (here we use the above-mentioned characterization of \pi as a zero of the sine function), Claim 2 follows.
Conclusion: Since f(x) > 0 and \sin x > 0 for 0 (because \pi is the
smallest positive zero of the sine function), Claims 1 and 2 show that F(0) + F(\pi) is a
positive integer. Since 0 \leq x(a - bx) \leq \pi a and 0 \leq \sin x \leq 1 for 0 \leq x \leq \pi, we have, by the original definition of f, :\int_0^\pi f(x)\sin(x)\,dx\le\pi\frac{(\pi a)^n}{n!} which is smaller than 1 for large n, hence F(0) + F(\pi) for these n, by Claim 2. This is impossible for the positive integer F(0) + F(\pi). This shows that the original assumption that \pi is rational leads to a contradiction, which concludes the proof. The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula :\int_0^\pi f(x)\sin(x)\,dx = \sum_{j=0}^n (-1)^j \left (f^{(2j)}(\pi)+f^{(2j)}(0)\right )+(-1)^{n+1}\int_0^\pi f^{(2n+2)}(x)\sin(x)\,dx, which is obtained by 2n + 2
integrations by parts. Claim 2 essentially establishes this formula, where the use of F hides the iterated integration by parts. The last integral vanishes because f^{(2n+2)} is the zero polynomial. Claim 1 shows that the remaining sum is an integer. Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight. In fact, :\begin{align} J_n(x)&=x^{2n+1}\int_{-1}^1(1-z^2)^n\cos(xz)\,dz\\ &=\int_{-1}^1\left (x^2-(xz)^2\right )^nx\cos(xz)\,dz. \end{align} Therefore, the
substitution xz = y turns this integral into :\int_{-x}^x(x^2-y^2)^n\cos(y)\,dy. In particular, :\begin{align} J_n\left(\frac\pi2\right)&=\int_{-\pi/2}^{\pi/2}\left(\frac{\pi^2}4-y^2\right)^n\cos(y)\,dy\\[5pt] &=\int_0^\pi\left(\frac{\pi^2}4-\left(y-\frac\pi2\right)^2\right)^n\cos\left(y-\frac\pi2\right)\,dy\\[5pt] &=\int_0^\pi y^n(\pi-y)^n\sin(y)\,dy\\[5pt] &=\frac{n!}{b^n}\int_0^\pi f(x)\sin(x)\,dx. \end{align} Another connection between the proofs lies in the fact that Hermite already mentions that if f is a polynomial function and :F=f-f^{(2)}+f^{(4)}\mp\cdots, then :\int f(x)\sin(x)\,dx=F'(x)\sin(x)-F(x)\cos(x)+C, from which it follows that :\int_0^\pi f(x)\sin(x)\,dx=F(\pi)+F(0). == Bourbaki's proof ==