Irrationality of pi
We’re getting close! Last time, we defined a new function 
and showed that 
and 
are both integers, and that 
. So, consider the following:
![$ \begin{align*} &\frac{d}{dx} [ F'(x) \sin x - F(x) \cos x ] \\ &= F^{\prime\prime}(x)\sin x + F'(x) \cos x \\ & \qquad - F'(x) \cos x + F(x) \sin x \\ &= F^{\prime\prime}(x) \sin x + F(x) \sin x \\ &= [F^{\prime\prime}(x) + F(x)]\sin x \\ &= f(x) \sin x. \end{align*} $](http://wso.williams.edu/~byorgey/wordpress/wp-content/plugins/latexrender/pictures/e2aa0056c82331c22cdc81cddc6d8680.gif "$ \begin{align*} &\frac{d}{dx} [ F'(x) \sin x - F(x) \cos x ] \\ &= F^{\prime\prime}(x)\sin x + F'(x) \cos x \\ & \qquad - F'(x) \cos x + F(x) \sin x \\ &= F^{\prime\prime}(x) \sin x + F(x) \sin x \\ &= [F^{\prime\prime}(x) + F(x)]\sin x \\ &= f(x) \sin x. \end{align*} $")
The first step uses the product rule for differentiation (recalling that 
and 
); the last step is what we showed last time. Now we see the point of defining : it’s just so that we have a convenient way to talk about the antiderivative of 
. We could just do everything directly in terms of alternating sums of derivatives of 
… but it’s much clearer this way, don’t you agree?
Now that we know the antiderivative of 
, we can use the Fundamental Theorem of Calculus to compute the following integral:
![$ \begin{align*} &\int_0^\pi f(x)\sin x dx \\ &= \left[ F'(x) \sin x - F(x) \cos x \right]_0^\pi \\ &= F'(\pi) \sin \pi - F(\pi) \cos \pi \\ & \qquad - F'(0) \sin 0 + F(0) \cos 0 \\ &= F(\pi) + F(0). \end{align*} $](http://wso.williams.edu/~byorgey/wordpress/wp-content/plugins/latexrender/pictures/dc4b4ef4ced9e046e140679279400b9b.gif "$ \begin{align*} &\int_0^\pi f(x)\sin x dx \\ &= \left[ F'(x) \sin x - F(x) \cos x \right]_0^\pi \\ &= F'(\pi) \sin \pi - F(\pi) \cos \pi \\ & \qquad - F'(0) \sin 0 + F(0) \cos 0 \\ &= F(\pi) + F(0). \end{align*} $")
Note that the value of this integral is an integer, since both and are integers. But next time we’ll show that it is also strictly between 0 and 1 (for a suitable choice of 
), which is clearly nonsense!
