4th Putnam 1941

Problem B6

f is a real valued function on [0, 1], continuous on (0, 1). Prove that ∫x=0x=1y=xy=1z=xz=y f(x) f(y) f(z) dz dy dx = 1/6 ( ∫x=0x=1 f(x) dx )3.



Let Sxyz be the points in the cube for which x ≤ y ≤ z, let Syxz be the points for which y ≤ x ≤ z and so on. Then the union of the six sets is the cube and the intersection of any two has measure zero. Also by changing the variables of integration we see that the integral of f(x) f(y) f(z) over each set is the same. Hence the integral over Sxzy is 1/6 of the integral over the cube. But the integral over Sxzy is just ∫x=0x=1y=xy=1z=xz=y f(x) f(y) f(z) dz dy dx and the integral over the cube is ( ∫x=0x=1 f(x) dx )3.



4th Putnam 1941

© John Scholes
5 Mar 2002