It goes (sort of) like this: let $q$ be any integer.

Let $A_n=q^n/n!{\int }_0^{\pi }{x}^n(\pi -x{)}^n\sin (x)\mathrm{dx}$

Then show, by integrating $A_n$ by parts twice, that $A_{n+2}=(4n+6)q{A}_{n+1}-(q\pi {)}^2{A}_n$. This is quite fiddly, as it is easy to get very confused with the sheer volume of algebra, but it doesn't need you to do anything particularly clever.

Now for the important bit. Suppose $\pi$ is rational; then we can write $\pi =p/q$, so that both $q$ and $q\pi$ are integers.

Then if we use this $q$ in the recurrence relation above, we see that $A_{n+2}$ is a sum of integer multiples of $A_{n+1}$ and $A_n$. So if these are integers, $A_{n+2}$ is also an integer. We can check the values of $A_0$, which is 2, and $A_1$, which is $4q$. So, if our $q$ exists, then $A_0$, $A_1$, and hence $A_2$ and all the other $A$'s are integers.

Now we are almost there, although it may not look like it. How big is $A_n$? Since $\sin (x)\ge 0$ for $x=0..\pi$, $A_n$ is positive for all $n$.

However, for $x$ between 0 and $\pi$, $x^n\le p^n$,

$(p-x{)}^n\le p^n$,

and $\sin (x)\le 1$.

So $A_n$ is at most $q^n/n!{\int }_0^{\pi }{\pi }^{2n}\mathrm{dx}=\pi (q{\pi }^2{)}^n/n!$.

Can you see that as we make n larger and larger, this will inevitably get smaller and smaller? In other words, it tends to 0 as $n$ gets big.

So, we know three things about $A_n$:

• it's getting smaller and smaller
• it's never zero
• it's an integer

So we have, at last, reached a contradiction. Thus our original assumption that $q$ existed must have been false; we can't write $\pi =p/q$, i.e. it is irrational. QED.