Assume for contradiction that

π = a / b

with positive integers

a, b

F_{n} (x) = \frac{b ^{n}}{n !} x^{n} (π - x)^{n} = \frac{1}{n !} x^{n} (a - b x)^{n}

and

I_{n} = \int_{0}^{π} F_{n} (x) sin x d x .

x^{n + j}

F_{n}

\frac{( - 1 ) ^{j}}{n !} (j n) a^{n - j} b^{j} .

Consequently every derivative

F_{n}^{(k)} (0)

is an integer: it is zero unless

n \leq k \leq 2 n

, and in that range the factor

k! / n!

clears the denominator. Since

F_{n} (π - x) = F_{n} (x)

, every

F_{n}^{(k)} (π)

is also an integer.

Repeated integration by parts, ending when derivatives above degree

2 n

vanish, gives

I_{n} = \sum_{j = 0}^{n} (- 1)^{j} (F_{n}^{(2 j)} (0) + F_{n}^{(2 j)} (π)) .

Thus

I_{n}

is an integer. It is strictly positive because

F_{n} (x) sin x > 0

for

0 < x < π

On the other hand,

x (π - x) \leq π^{2} /4

, so

0 < I_{n} \leq \frac{π}{n !} (\frac{b π ^{2}}{4})^{n} .

The factorial-over-power series implies that the right-hand side tends to zero. For all sufficiently large

n

this says that the positive integer

I_{n}

is less than one, a contradiction. Therefore the Irrationality of pi is established:

π \in / Q .

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 Incoming links: Irrationality of pi