X is the unit n-cube, [0, 1]^{n}. Let k_{n} = ∫_{X} cos^{2}( π(x_{1} + x_{2} + ... + x_{n})/(2n) ) dx_{1} ... dx_{n}. What is lim_{n→∞} k_{n} ?

**Solution**

Let y_{i} = 1 - x_{i} and change variables from x_{i} to y_{i}. The sum (x_{1} + x_{2} + ... + x_{n})/(2n) becomes 1/2 - (y_{1} + y_{2} + ... + y_{n})/(2n), so the integrand becomes sin^{2}( π(y_{1} + y_{2} + ... + y_{n})/(2n) ). So the integral is identical to the original except that cos has been changed to sin. Thus we can add it to the original to get 2k_{n} = ∫_{X} dx_{1} dx_{2} ... dx_{n} = 1. So k_{n} = 1/2 for all n.

© John Scholes

jscholes@kalva.demon.co.uk

25 Jan 2002