If I asked the obvious question because of my lack of knowledge, then I am sorry! Assuming $\pi$ is a rational number, ie $\pi = \frac{a}{b} $ and a, b is an irreducible integer,ie gcd(a,b)=1.Let:$$f{(x)}=\frac{x^n(a-bx)^n}{n!},n\in\mathbb{N^+}$$
- Prove:$\int_0^{\pi}f{(x)}\sin{x}\,dx$ is a integer.
- It follows that $\pi$ can not be a rational number Using the segmentation method.
I know:$$\int_0^{\pi}f{(x)}\sin{x}\,dx$$ $$=-\cos{x}f{(x)}|_0^{\pi}+\int_0^{\pi}\cos{x}f^{'}(x)\,dx$$ $$=2f{(\pi)}+\sin{x}f^{'}{(x)}|_0^{\pi}-\int_0^{\pi}\sin{x}f^{"}{(x)}\,dx$$ $$=\frac{2{\pi}^n{(a-bx)}^n}{n!}-\int_0^{\pi}\sin{x}f^{"}{(x)}\,dx$$But $\int_0^{\pi}\sin{x}f^{"}{(x)}\,dx$ is too complex, and if it continues, will be more and more complex.And I know:$$f{(\frac{a}{b}-x)}=f{(x)}$$I would like to know whether there is a simple way.Could anyone help me?Thank you!
Niven's Original Paper is pretty clear and simple. In the following I'll give you some hints to better understand that paper, with its notation:
(1) Be sure you can prove $\;f^{(k)}(0)\;,\;\;f^{(k)}(\pi)\;,\;\;\;k\ge0\;$ , are integers. Observe that both $\;0,\,\pi\;$ are roots of the polynomial $\;f\;$ .
(2) You need to show $\;F''+F=f\;$ , but this in fact follows from the fact that $\;\deg f=2n\;$ ...
(3) Now you can deduce $\;F(0),\,F(\pi)\;$ are integers
(4) Finally, it must be clear that $\;x\in (0,\pi]\implies f(x)\sin x>0\;$ , and the other inequality you need follows from the fact that $\;\cfrac\pi2\;$ is a maximum.
And don't forget that the assumption is $\;\pi=\cfrac ab\;,\;\;0<a,\,b\in\Bbb Z\;$ .