Random integrals

An enlightening discussion about pi and tau.

Random integrals

by SpikedMath » Thu Jul 07, 2011 2:59 pm

In the Pi Manifesto we had some integrals that converged to [tex]\pi[/tex]:

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

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

by qmod » Wed Jul 13, 2011 2:39 am

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

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

[tex]\frac{x}{1!}, \frac{x^2}{2!}, \frac{x^3}{3!}, \frac{x^4}{4!}[/tex]
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Re: Random integrals

by Chris Park » Thu Jul 14, 2011 5:05 pm

qmod, when you say integrals "bring out" 1/2, are you just referring to power rule? (i.e., the antiderivative of [tex]u[/tex] is [tex]\frac{1}{2} u^2[/tex])? 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 [tex]1/2 \tau r^2[/tex] could be seen as [tex]1/2! \tau r^2[/tex], but that doesn't create a pattern with [tex]4/3! \tau r^3[/tex] 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. [tex]pi/3[/tex] and [tex]\tau/3![/tex] 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.
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Re: Random integrals

by Kodegadulo » Mon Jul 02, 2012 12:31 pm

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 [tex]\tau[/tex]. If you pollute all the formulas with powers of [tex]2\pi[/tex] and exploit those powers of 2 to cancel out factors in the even-dimensions, the similarities between the even and odd dimensions get obscured.
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Re: Random integrals

by Kodegadulo » Mon Jul 02, 2012 12:59 pm

SpikedMath wrote:In the Pi Manifesto we had some integrals that converged to [tex]\pi[/tex]:

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

It's intriguing to me that each of those are exactly equal to [tex]\pi[/tex]! Are there any nice examples that are exactly equal to [tex]\tau[/tex]? (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, [tex]\theta_1[/tex] and [tex]\theta_2[/tex], that are half a turn apart: [tex]\theta_2 - \theta_1 = \tau/2 = \pi[/tex]
  • Pick some trig formula [tex]f\left(\theta\right)[/tex]
  • Evaluate the formula at those two angles, get values [tex]f\left(\theta_1\right) = f_1[/tex] and [tex]f\left(\theta_2\right) =f_2[/tex].
  • Find the inverse of [tex]f[/tex], call it [tex]g\left(x\right)[/tex], such that [tex]g\left(f_1\right) = \theta_1[/tex] and [tex]g\left(f_2\right) =\theta_2[/tex]
  • Take the derivative of your inverse formula: [tex]g'\left(\theta\right)[/tex].
  • Set up an integral [tex]\int_{f_1}^{f_2} g'\left(x\right)[/tex]
  • That "magically" evaluates to [tex]g\left(f_2\right) - g\left(f_1\right) = \theta_2 - \theta_1 = \tau/2 = \pi[/tex].

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

by rdococ » Fri Oct 11, 2013 5:26 pm

When you get the integral where [tex]a = -1[/tex] and [tex]b = 1[/tex], the signed and unsigned parts 'cancel out' on an other-wise [tex]\tau[/tex] friendly integral.

Also, integrals bring out a factor of a half when we take smaller ranges, and signed functions also bring factors when we take any range. It only makes sense for us to describe that factor, instead of hiding it in a [tex]\pi[/tex].

So, technically, [tex]\tau[/tex] wins without its own examples when it comes to integrals.
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Re: Random integrals

by 1=0 » Sun Mar 19, 2017 11:48 am

The integral of any trig function from 0 to tau is zero.
There's also this one: https://wikimedia.org/api/rest_v1/media ... 09e44dc425
And the very important Fourier transform, which is simpler with tau, uses integrals.
But of course, oviously the most important example is integrals of the hydrodynamic equations corresponding to vortex motion: https://youtu.be/Bcr9-93wXng
Look at that equation behind him at 2:23."That's a triple integral right there for crying out loud. What more do you want?"
But seriously, I think the Fourier transform is a good enough example (and certainly important enough) to show that pi doesn't always win when it comes to integrals.

Also, the integral of 1/sqrt(1-x^2) only gets you half of the circumference of a unit circle because sqrt(1-x^2) is only a semicircle. To get a full circle, you have to graph both sqrt(1-x^2) and -sqrt(1-x^2). Sure enough, the area between 1/sqrt(1-x^2) and -1/sqrt(1-x^2) is tau, the circumference of an entire unit circle.

I also seem to have found a win for eta:
[tex]\int_{-\infty}^\infty \frac{\sin x \cos x}{x}\,dx = \eta.[/tex]
Last edited by 1=0 on Sat Apr 22, 2017 10:08 am, edited 3 times in total.
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Re: Random integrals

by 1=0 » Sun Mar 19, 2017 11:59 am

Kodegadulo wrote:
SpikedMath wrote:In the Pi Manifesto we had some integrals that converged to [tex]\pi[/tex]:

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

It's intriguing to me that each of those are exactly equal to [tex]\pi[/tex]! Are there any nice examples that are exactly equal to [tex]\tau[/tex]? (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, [tex]\theta_1[/tex] and [tex]\theta_2[/tex], that are half a turn apart: [tex]\theta_2 - \theta_1 = \tau/2 = \pi[/tex]
  • Pick some trig formula [tex]f\left(\theta\right)[/tex]
  • Evaluate the formula at those two angles, get values [tex]f\left(\theta_1\right) = f_1[/tex] and [tex]f\left(\theta_2\right) =f_2[/tex].
  • Find the inverse of [tex]f[/tex], call it [tex]g\left(x\right)[/tex], such that [tex]g\left(f_1\right) = \theta_1[/tex] and [tex]g\left(f_2\right) =\theta_2[/tex]
  • Take the derivative of your inverse formula: [tex]g'\left(\theta\right)[/tex].
  • Set up an integral [tex]\int_{f_1}^{f_2} g'\left(x\right)[/tex]
  • That "magically" evaluates to [tex]g\left(f_2\right) - g\left(f_1\right) = \theta_2 - \theta_1 = \tau/2 = \pi[/tex].

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


I looked at the indefinite integrals, and you seem to be correct. The indefinite integral of sech x is arctan(sinh x)+C, and the indefinite integral of 1/(1+x^2) is arctan x + C. This of course means that the integrals over all space will be the range of arctan, which is the period of tangent, which is pi, but only because tangent's values repeat halfway around the circle.

I still haven't been able to find out why the integral of (1-cos x)/x^2 is pi, but I suspect it's something similar. I also have to wonder why the Pi Manifesto calls it a "well known" formula if I can't find it anywhere.
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