## Random integrals

An enlightening discussion about pi and tau.

### Random integrals

In the Pi Manifesto we had some integrals that converged to $\pi$:

$\int_{-\infty}^\infty \operatorname{sech}(x)\,dx = \pi.$
$\int_{-1}^1 \frac{1}{\sqrt{1-x^2}}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{1-\cos x}{x^2}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{\sin x}{x}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{\sin^2 x}{x^2}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{1}{1+x^2}\,dx = \pi.$

It's intriguing to me that each of those are exactly equal to $\pi$! Are there any nice examples that are exactly equal to $\tau$? (That is, without multiplying the above ones by 2 or sticking 1/2's throughout somewhere).
Math - It's in you to give.

SpikedMath

Posts: 133
Joined: Mon Feb 07, 2011 1:31 am

### Re: Random integrals

I think the idea is integrals automatically bring half most of the time so $\tau/2$ is only natural.

I think if you were to double integrate, you might get pi/3 instead of tau/3! Imagine volume of sphere
$\frac{4}{3} \pi r^3 = \frac{4}{3!} \tau r^3$.
(If I had taken cone that 4 factor wouldn't be there too)

$\frac{x}{1!}, \frac{x^2}{2!}, \frac{x^3}{3!}, \frac{x^4}{4!}$
qmod
Kindergarten

Posts: 3
Joined: Mon Feb 14, 2011 9:49 am

### Re: Random integrals

qmod, when you say integrals "bring out" 1/2, are you just referring to power rule? (i.e., the antiderivative of $u$ is $\frac{1}{2} u^2$)? Otherwise, I don't really see what you mean.

I don't really follow what you're trying to say with the factorial expressions there. Are you saying that factorials belong in the denominator because of power rule? I get that $1/2 \tau r^2$ could be seen as $1/2! \tau r^2$, but that doesn't create a pattern with $4/3! \tau r^3$ because of that 4.

And yes, the equation for the volume of a cone fits the pattern (without the extra 4), but how on earth is a cone the 3-dimensional analog to a circle?

And I don't follow your sentence about double integration, either. $pi/3$ and $\tau/3!$ are the same thing, right? And how would you double integrate those functions of one variable? Do you mean take the antiderivative twice?

Forgive my many questions in one post, but I couldn't figure out your meaning is all.
Chris Park
Kindergarten

Posts: 8
Joined: Sat Jul 09, 2011 1:25 am

### Re: Random integrals

The actual pattern for n-sphere surface areas and volumes is all spelled out in my post in another thread. It involves double-factorials (n!!) which are like factorials except the decrement is 2 rather than 1. The even and odd dimensions are completely separate (evens build on evens, odds build on odds), but other than different starting coefficients for the lowest cases, the pattern for both is the same. That is, as long as you derive them all using powers of $\tau$. If you pollute all the formulas with powers of $2\pi$ and exploit those powers of 2 to cancel out factors in the even-dimensions, the similarities between the even and odd dimensions get obscured.
Kindergarten

Posts: 6
Joined: Sun Jun 10, 2012 3:04 am

### Re: Random integrals

SpikedMath wrote:In the Pi Manifesto we had some integrals that converged to $\pi$:

$\int_{-\infty}^\infty \operatorname{sech}(x)\,dx = \pi.$
$\int_{-1}^1 \frac{1}{\sqrt{1-x^2}}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{1-\cos x}{x^2}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{\sin x}{x}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{\sin^2 x}{x^2}\,dx = \pi.$
$\int_{-\infty}^\infty \frac{1}{1+x^2}\,dx = \pi.$

It's intriguing to me that each of those are exactly equal to $\pi$! Are there any nice examples that are exactly equal to $\tau$? (That is, without multiplying the above ones by 2 or sticking 1/2's throughout somewhere).

I haven't investigated all of these, but likely what these amount to is the following procedure:
• Pick two angles, $\theta_1$ and $\theta_2$, that are half a turn apart: $\theta_2 - \theta_1 = \tau/2 = \pi$
• Pick some trig formula $f\left(\theta\right)$
• Evaluate the formula at those two angles, get values $f\left(\theta_1\right) = f_1$ and $f\left(\theta_2\right) =f_2$.
• Find the inverse of $f$, call it $g\left(x\right)$, such that $g\left(f_1\right) = \theta_1$ and $g\left(f_2\right) =\theta_2$
• Take the derivative of your inverse formula: $g'\left(\theta\right)$.
• Set up an integral $\int_{f_1}^{f_2} g'\left(x\right)$
• That "magically" evaluates to $g\left(f_2\right) - g\left(f_1\right) = \theta_2 - \theta_1 = \tau/2 = \pi$.

But this is essentially gaming the system.When I have some time, I'll see if I can hunt down what $f$ and $g$ is for each of these. For instance, the second one looks to me like $f\left(\theta\right) = \sin \theta$ and $g\left(x\right) = \arcsin x$, where $\sin\left(\frac{\tau}{4}\right) = 1$ and $\sin\left(-\frac{\tau}{4}\right) = -1$.