Irrationality of pi: the impossible integral

We’re getting close! Last time, we defined a new function $F(x)$ and showed that $F(0)$ and $F(\pi)$ are both integers, and that $F^{\prime\prime}(x) + F(x) = f(x)$ . So, consider the following:

$\begin{array}{cl} &\frac{d}{dx} [ F'(x) \sin x - F(x) \cos x ] \\ = & F^{\prime\prime}(x)\sin x + F'(x) \cos x \\ & \qquad - F'(x) \cos x + F(x) \sin x \\ = & F^{\prime\prime}(x) \sin x + F(x) \sin x \\ = & [F^{\prime\prime}(x) + F(x)]\sin x \\ = & f(x) \sin x. \end{array}$

The first step uses the product rule for differentiation (recalling that $\frac{d}{dx}\sin x = \cos x$ and $\frac{d}{dx}\cos x = - \sin x$ ); the last step is what we showed last time. Now we see the point of defining $F(x)$ : it’s just so that we have a convenient way to talk about the antiderivative of $f(x) \sin x$ . We could just do everything directly in terms of alternating sums of derivatives of $f(x)$ … but it’s much clearer this way, don’t you agree?

Now that we know the antiderivative of $f(x)\sin x$ , we can use the Fundamental Theorem of Calculus to compute the following integral:

$\begin{array}{cl} & \int_0^\pi f(x)\sin x dx \\ = & \left[ F'(x) \sin x - F(x) \cos x \right]_0^\pi \\ = & F'(\pi) \sin \pi - F(\pi) \cos \pi \\ & \qquad - F'(0) \sin 0 + F(0) \cos 0 \\ = & F(\pi) + F(0). \end{array}$

Note that the value of this integral is an integer, since both $F(\pi)$ and $F(0)$ are integers. But next time we’ll show that it is also strictly between 0 and 1 (for a suitable choice of $n$ ), which is clearly nonsense!

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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4 Responses to Irrationality of pi: the impossible integral

Jack says:

February 6, 2010 at 2:06 pm

Great! I see how to finish it off now, and I see why given the differential equation and being told to look at F and f you would see that relation, but just introducing F in the first place… well, I guess I just have to accept this proof is awesome. =D

You asked for topic areas: Have you ever seen Euler’s proof that the number of partitions of a number into odd numbers equals that of a number into distinct numbers? That is a good one. Other than that, some Galois theory would be nice?
Brent says:

February 6, 2010 at 4:14 pm

Jack: I have seen that partition proof, that might work well! I think Galois theory might be a little too deep for the intended readership of this blog.
Dave says:

February 9, 2010 at 7:01 pm

Very cool! Been a long time since I’ve used the chain rule. Thank goodness for Google! I’m still with ya. Looking forward to the final installment.
Dave says:

February 9, 2010 at 7:02 pm

Sorry. That last post should have said, “product rule.”

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