Use the method of stationary phase to estimate, by hand, the average value of \[ \sin^{100}(x). \] Do not use a calculator. The exact answer is about \(1/12\).
Solution
The stationary-phase/Laplace estimate for the average over one period is \[ \boxed{ \operatorname{avg}(\sin^{100}x) \approx\frac1{\sqrt{50\pi}}.} \] This is roughly \(0.08\), hence close to \(1/12\).
1. By symmetry, the average can be written\[ A =\frac2\pi\int_0^{\pi/2}\sin^{100}x\,dx. \]
The function \(\sin^{100}x\) has period \(\pi\) and is symmetric in each quadrant. Hence \[ \int_0^{2\pi}\sin^{100}x\,dx =4\int_0^{\pi/2}\sin^{100}x\,dx. \] Dividing by the period length \(2\pi\) gives \[ A =\frac1{2\pi}\int_0^{2\pi}\sin^{100}x\,dx =\frac2\pi\int_0^{\pi/2}\sin^{100}x\,dx. \]
2. Center the integral at the maximum \(x=\pi/2\) by setting\(t=\pi/2-x\). Then\[ A=\frac2\pi\int_0^{\pi/2}\cos^{100}t\,dt. \]
Under the substitution \(t=\pi/2-x\), \[ \sin x=\cos t, \] and reversing the limits yields the displayed formula.
3. Near \(t=0\), the exponent has the quadratic approximation\[ 100\log(\cos t)=-50t^2+O(t^4). \]
The Taylor expansion at the stationary point is \[ \cos t=1-\frac{t^2}{2}+O(t^4). \] Since \(\log(1+u)=u+O(u^2)\), \[ \log(\cos t) =-\frac{t^2}{2}+O(t^4). \] Multiplying by \(100\) gives \[ 100\log(\cos t) =-50t^2+O(100t^4). \] Thus the peak has Gaussian width on the scale \(t\asymp100^{-1/2}\).
4. Replacing the peak by its Gaussian model gives\[ \int_0^{\pi/2}\cos^{100}t\,dt \approx\int_0^\infty e^{-50t^2}\,dt =\frac12\sqrt{\frac\pi{50}}. \]
Away from a small neighborhood of \(0\), \(|\cos t|<1\) by a fixed amount, so raising to the \(100\)-th power suppresses that part of the integral. Inside the peak, 3 replaces \(\cos^{100}t\) by \(e^{-50t^2}\) to leading order. Extending the short Gaussian peak to the positive half-line changes only the lower-order tail. The standard Gaussian integral gives \[ \int_0^\infty e^{-50t^2}\,dt =\frac12\sqrt{\frac\pi{50}}. \]
5. Therefore\[ A\approx\frac1{\sqrt{50\pi}}. \]
Substituting 4 into 2, \[ A \approx\frac2\pi\cdot\frac12\sqrt{\frac\pi{50}} =\frac1{\sqrt{50\pi}}. \] For a hand estimate, use \(3<\pi<22/7\). Then \[ \frac1{\sqrt{1100/7}} <\frac1{\sqrt{50\pi}} <\frac1{\sqrt{150}}. \] Both endpoints are about eight hundredths; in particular this predicts a value close to \[ \frac1{12}\approx0.083. \] This agrees with the scale stated in the problem.