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 these ads
This entry was posted in famous numbers, proof and tagged , , , , , . Bookmark the permalink.

4 Responses to Irrationality of pi: the impossible integral

  1. Jack says:

    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?

  2. Brent says:

    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.

  3. Dave says:

    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.

  4. Dave says:

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

Comments are closed.