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].