Let a(r) be the number of lattice points inside the circle center the origin, radius r. Let k = 1 + e-1 + e-4 + ... + exp(-n2) + ... . Express ∫U a(√(x2 + y2) ) exp( -(x2+y2) ) dx dy as a polynomial in k, where U represents the entire plane.
Solution
Answer: π(4k2 -4k + 1).
a(r) is constant except at certain values √n where it increases. Let an be the increase at √n, and a0 = 1, so that a(r) = a0 + a1 + ... + an on the interval (√n, √(n+1) ).
In the integral given, change variables to r, θ. The Jacobian is r, so we get 2π ∫0∞ r a(r) exp( - r2) dr = 2π ∑0∞ ∫√n√(n+1) r a(r) exp( - r2) dr = 2π ∑0∞ ∫√n√(n+1) r (a0 + a1 + ... + an) exp( - r2) dr = π ( a0(1 - e-1) + (a0 + a1)(e-1 - e-2) + (a0 + a1 + a2)(e-2 - e-3) + ... ) = π(a0 + a1e-1 + a2e-2 + ... ).
Let p(x) = 1 + x + x4 + x9 + x16 + ... . Then the coefficient of xn in p(x)2 is the number of lattice points in the first quadrant on the circle radius √n centre the origin. The total number of lattice points on the circle is not 4p(x)2 because that double-counts the points on the axes, it is 4p(x)2 - 4p(x) + 1 (the final 1 is to put back the point at the origin). Hence a0 + a1e-1 + a2e-2 + ... = 4p(1/e)2 -4p(1/e) + 1 = 4k2 -4k + 1 and the integral is π(4k2 -4k + 1).
© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001