Computational Complexity of the Divisor summatory function
Based on a "folkore intuition" for a related problem, one might expect this to be at least as hard as factoring: 

The folkore intuition suggests that computing the number of prime factors, or even computing essentially any useful information about the prime factorization, is as hard as factoring itself (see Tao's answer to this related question). $d(n)$ gives you exactly such information, and is essentially equivalent (from the computational perspective) to computing $D(n)$, since $d(n) = D(n) - D(n-1)$.

For example, if $n$ is squarefree, then $d(n)$ is equal to $2^{f}$ where $f$ is the number of prime factors. For general $n$, $d(n)$ is $\prod_{p} (m_p + 1)$ where $m_p$ is the number of times the prime $p$ divides $n$ (that is, $n = \prod_{p} p^{m_p}$ is the prime factorization of $n$).