Problem for understanding Dirichlet Parabola. Let $\tau(n)$ the number of divisor of $n$. I have to prove that for all $1\leq y\leq x$ $$\sum_{n\leq x}\tau(n)=\sum_{d\leq y}\sum_{m\leq \frac{x}{d}}1+\sum_{m\leq \frac{x}{y}}\sum_{d\leq \frac{x}{m}}1-\sum_{d\leq y}\sum_{m\leq \frac{x}{y}}1.$$
So, $\tau(n)=\sum_{d\mid n}1$. There $$\sum_{n\leq x}\tau(n)=\sum_{n\leq x}\sum_{dm=n}1$$
So $$\{(n,d)\mid n\leq x, \exists k: dk=n\}=\{(d,k)\mid dk\leq x\}=\{(d,k)\mid d\leq y, k\leq \frac{x}{d}\}\cup\{(d,k)\mid y<d\leq x, k\leq \frac{x}{d}\},$$
but I can't do better. Any idea ?
 A: You are counting the number of pairs $(d,m)$ of positive integers such that $d\cdot m \leqslant x$. For a fixed $d$, a pair $(d,m)$ belongs to this set if and only if $0 < m \leqslant \dfrac{x}{d}$, so there are $\biggl\lfloor\dfrac{x}{d}\biggr\rfloor$ pairs with first component $d$. Symmetrically, there are $\biggl\lfloor\dfrac{x}{m}\biggr\rfloor$ pairs with second component $m$.
Given a positive $y$, you can split those pairs into two groups, those with $d \leqslant y$ (call the set of these $A$) and those with $d > y$ (call that set $B$). Counting the pairs in the first group gives
$$\sum_{d\leqslant y} \biggl\lfloor\frac{x}{d}\biggr\rfloor = \sum_{d\leqslant y} \sum_{m \leqslant \frac{x}{d}} 1.$$
Similarly, we can split the set of pairs into two subsets according to whether $m \leqslant \dfrac{x}{y}$ (those make up set $C$) or $m > \dfrac{x}{y}$ (and $D$). Counting the pairs in $C$ gives
$$\sum_{m \leqslant \frac{x}{y}}\biggl\lfloor\frac{x}{m}\biggr\rfloor = \sum_{m \leqslant \frac{x}{y}} \sum_{d \leqslant \frac{x}{m}} 1.$$
Now every pair $(d,m)$ with $d\cdot m \leqslant x$ belongs to $A$ or $C$ - if $(d,m) \notin A$, so $d > y$, then $m \leqslant \dfrac{x}{d} < \dfrac{x}{y}$, hence $(d,m) \in C$ - so we have counted all of the pairs we wanted to count. But $A$ and $C$ are not disjoint (unless $y < 1$ or $y > x$, when at least one of $A$ and $C$ is empty). We have counted the pairs belonging to $A\cap C$ twice, so we need to subtract the number of pairs in $A\cap C$ to get the correct count. But $(d,m) \in A \cap C$ if and only if $d \leqslant y$ and $m \leqslant \dfrac{x}{y}$, so the number of pairs in $A\cap C$ is
$$\lfloor y\rfloor \cdot \biggl\lfloor \frac{x}{y}\biggr\rfloor = \sum_{d\leqslant y} \sum_{m \leqslant \frac{x}{y}} 1.$$
Altogether,
\begin{align}
\sum_{n \leqslant x} \tau(n)
&= \sum_{\substack{d,m \\ d\cdot m \leqslant x}} 1 \\
&= \underbrace{\sum_{d\leqslant y}\sum_{m\leqslant \frac{x}{d}} 1}_{\lvert A\rvert} + \underbrace{\sum_{m \leqslant \frac{x}{y}}\sum_{d \leqslant \frac{x}{m}} 1}_{\lvert C\rvert} - \underbrace{\sum_{d\leqslant y}\sum_{m\leqslant \frac{x}{y}} 1}_{\lvert A\cap C\rvert}.
\end{align}
