We’re getting close! Last time, we defined a new function and showed that and are both integers, and that . So, consider the following:
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:
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!