  If $P(x)$ is the error term in the circle problem, then it is proved that
$$\int_0^\infty P^2(x)e^{-x/T}dx = {1\over4}({T\over\pi})^{3/2}
\sum_{n=1}^\infty r^2(n)n^{-3/2} - T + O_\epsilon(T^{2/3+\epsilon}), $$
improving the author's earlier exponent 5/6. The new bound is obtained by using
results of F. Chamizo on the correlated sum $\sum_{n\le x}r(n)r(n+h)$, where
$r(n)$ is the number of representations of $n$ as a sum of two squares.
