The number is not rational.

Proof

For any fixed real number $q$ , and any natural number $n$ , let $A_{n} = \frac{q^{n}}{n!} \int_{0}^{π} [x (π - x)]^{n} sin (x) d x$ where $n!$ is the factorial of $n$ , $\int$ is the definite_integral, and $sin$ is the sin_function.

Preparatory work

Exercise: $A_{n} = (4 n - 2) q A_{n - 1} - (q π)^{2} A_{n - 2}$ .

Show solution

We use integration_by_parts.

show this

Now, $A_{0} = \int_{0}^{π} sin (x) d x = 2$ so $A_{0}$ is an integer.

Also $A_{1} = q \int_{0}^{π} x (π - x) sin (x) d x$ which by a simple calculation is $4 q$ .

Show calculation

Expand the integrand and then integrate by parts repeatedly: $\frac{A_{1}}{q} = \int_{0}^{π} x (π - x) sin (x) d x = π \int_{0}^{π} x sin (x) d x - \int_{0}^{π} x^{2} sin (x) d x$

The first integral term is $[- x cos (x)]_{0}^{π} + \int_{0}^{π} cos (x) d x = π$

The second integral term is $[- x^{2} cos (x)]_{0}^{π} + \int_{0}^{π} 2 x cos (x) d x$ which is $π^{2} + 2 ([x sin (x)]_{0}^{π} - \int_{0}^{π} sin (x) d x)$ which is $π^{2} - 4$

Therefore $\frac{A_{1}}{q} = π^{2} - (π^{2} - 4) = 4$

Therefore, if $q$ and $q π$ are integers, then so is $A_{n}$ inductively, because $(4 n - 2) q A_{n - 1}$ is an integer and $(q π)^{2} A_{n - 2}$ is an integer.

But also $A_{n} \to 0$ as $n \to \infty$ , because $\int_{0}^{π} [x (π - x)]^{n} sin (x) d x$ is in modulus at most $π \times max 0 \leq x \leq π [x (π - x)]^{n} sin (x) \leq π \times max 0 \leq x \leq π [x (π - x)]^{n} = π \times {[\frac{π^{2}}{4}]}^{n}$ and hence $| A_{n} | \leq \frac{1}{n!} {[\frac{π^{2} q}{4}]}^{n}$

For $n$ larger than $\frac{π^{2} q}{4}$ , this expression is getting smaller with $n$ , and moreover it gets smaller faster and faster as $n$ increases; so its limit is $0$ .

Formal treatment

We claim that $\frac{r^{n}}{n!} \to 0$ as $n \to \infty$ , for any $r > 0$ .

Indeed, we have $\frac{r^{n + 1} / (n + 1)!}{r^{n} / n!} = \frac{r}{n + 1}$ which, for $n > 2 r - 1$ , is less than $\frac{1}{2}$ . Therefore the ratio between successive terms is less than $\frac{1}{2}$ for sufficiently large $n$ , and so the sequence must shrink at least geometrically to $0$ .

Conclusion

Suppose (for contradiction) that $π$ is rational; then it is $\frac{p}{q}$ for some integers $p, q$ .

Now $q π$ is an integer (indeed, it is $p$ ), and $q$ is certainly an integer, so by what we showed above, $A_{n}$ is an integer for all $n$ .

But $A_{n} \to 0$ as $n \to \infty$ , so there is some $N$ for which $| A_{n} | < \frac{1}{2}$ for all $n > N$ ; hence for all sufficiently large $n$ , $A_{n}$ is $0$ . We already know that $A_{0} = 2$ and $A_{1} = 4 q$ , neither of which is $0$ ; so let $N$ be the first integer such that $A_{n} = 0$ for all $n \geq N$ , and we can already note that $N > 1$ .

Then $0 = A_{N + 1} = (4 N - 2) q A_{N} - (q π)^{2} A_{N - 1} = - (q π)^{2} A_{N - 1}$ whence $q = 0$ or $π = 0$ or $A_{N - 1} = 0$ .

Certainly $q \neq 0$ because $q$ is the denominator of a fraction; and $π \neq 0$ by whatever definition of $π$ we care to use. But also $A_{N - 1}$ is not $0$ because then $N - 1$ would be an integer $m$ such that $A_{n} = 0$ for all $n \geq m$ , and that contradicts the definition of $N$ as the least such integer.

We have obtained the required contradiction; so it must be the case that $π$ is irrational.