Find the average of sin100(x)\sin^{100} (x) in 5 minutes?

I read this quote attributed to VI Arnold.

“Who can’t calculate the average value of the one hundredth power of the sine function within five minutes, doesn’t understand mathematics – even if he studied supermanifolds, non-standard calculus or embedding theorems.”

EDIT Source is “A mathematical trivium” A book of 100 problems that university students “should be able to solve”. The statement asks for calculation within 10% accuracy.

So the average value over the entire domain should be the same as the average value over [0,π/2]


So here’s what I did:

First, this graph would be a train of highly sharp peaks. The integrand would assume values close to zero a up till before it sharply rises to 1.

So up till some ϵ[0,π/2] we will have sinxx and for the remaining π/2ϵ interval I could find the area of triangle with base π/2ϵ and height 1


I believe in principal it should be possible to find an ϵ such that the above expression yields the exact answer. So I try to approximate it, no good. Then I try mathematica and it is looking like there is no ϵ for which the value I am expecting is even close to the actual value. I plot the original and find that my approximation is hopeless.

Not to mention that my 5 minutes were over. So I admit I do not understand mathematics and humbly ask if someone could:

  1. Point out my mistake (Other than that ϵ is probably incomputable within 5 mins)
  2. How the hell is this done in 5 minutes?enter image description here

The picture below has the sin100x in blue (bottom) and my approximation of it plotted against ϵ (pink). Although there is no reason for them to be together, the upper graph has a minima quite above the exact value of the integral.

Just realized

Let u=cosx.



Since sinx=eixeix2i
and, for kZ, 2π0eikxdx={0k02πk=0,
we have 2π0sin100xdx=12100100k=0(100k)2π0eikx(1)100kei(100k)xdx=(10050)21002π, and the average value is (10050)2100.

Source : Link , Question Author : Community , Answer Author : Andrés E. Caicedo

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