Hyperbola Area and the Logarithm

The area under the curve $y = 1/x$ from $x=1$ to $x=n$ is $\ln n$.

The number of lattice points under $y = n/x$ grows like $n \ln n$.

These two facts are related by a single scaling argument. And the gap between the curve and the lattice points is where one of the deepest unsolved problems in mathematics lives.

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Area under the hyperbola

The curve $xy = n$, or equivalently $y = n/x$, is a hyperbola. The area under it from $x=1$ to $x=n$ (above $y=0$) is:

$\int_1^n \frac{n}{x}\, dx = n \ln n$

This is exact. The lattice point count $D(n) = \sum_{x=1}^{n} \lfloor n/x \rfloor$ approximates this area, because $\lfloor n/x \rfloor$ is the height of the column of lattice points at position $x$.

The approximation: $D(n) \approx n \ln n$.

How good is this approximation? That is the Dirichlet Divisor Problem — addressed in Dirichlet's Divisor Problem.

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Why $\ln n$ appears from $1/x$

$\int_1^n \frac{1}{x}\, dx = \ln n$

This is the defining property of the natural logarithm. The hyperbola $y = 1/x$ is the unique curve whose area grows logarithmically.

Discrete version. The harmonic sum $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ approximates $\ln n$:

$H_n = \ln n + \gamma + O\!\left(\frac{1}{n}\right)$

where $\gamma \approx 0.5772\ldots$ is the Euler-Mascheroni constant — the "gap" between the harmonic series and the logarithm.

This gap $\gamma$ appears because the harmonic sum $H_n$ is the area of unit-width rectangles of height $1/k$, while $\ln n$ is the smooth area. The rectangles overshoot by exactly $\gamma$ in the limit.

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The symmetry argument for area

The lattice point count using symmetry:

$D(n) = 2\sum_{x=1}^{m} \left\lfloor \frac{n}{x} \right\rfloor - m^2 \qquad m = \lfloor\sqrt{n}\rfloor$

Geometrically: count all lattice points in the region $xy \leq n$, $x,y \geq 1$.

Split at the line $x = y = \sqrt{n}$:

• The region $x \leq \sqrt{n}$ contains $\sum_{x=1}^{m} \lfloor n/x \rfloor$ points.
• The region $y \leq \sqrt{n}$ contains the same by symmetry.
• The square $x \leq \sqrt{n}$, $y \leq \sqrt{n}$ is counted twice — subtract $m^2$.

This mirrors how the area $n\ln n$ can be computed as twice the area of the region $x \leq \sqrt{n}$ minus the square of area $n$ (since $\sqrt{n} \times \sqrt{n} = n$):

$\int_1^{\sqrt{n}} \frac{n}{x}dx = n \cdot \frac{1}{2}\ln n$

Doubling and subtracting the square gives $n\ln n - n$ — matching the lattice point formula structurally.

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The error term

Define the error between the lattice count and the area approximation:

$\Delta(n) = D(n) - n\ln n - (2\gamma - 1)n$

Dirichlet proved $\Delta(n) = O(\sqrt{n})$ — the error grows no faster than $\sqrt{n}$.

The conjecture (still unproved): $\Delta(n) = O(n^{1/4+\varepsilon})$ for any $\varepsilon > 0$.

The best known result (as of 2025): $\Delta(n) = O(n^{131/416})$, due to Huxley. The true exponent is somewhere between $1/4$ and $131/416$.

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Where $\gamma$ comes from, concretely

$\sum_{x=1}^{n} \left\lfloor \frac{n}{x} \right\rfloor = n\ln n + (2\gamma - 1)n + O(\sqrt{n})$

The constant $2\gamma - 1$ appears because:

• The smooth approximation gives $n\ln n$
• The correction for replacing $\lfloor n/x \rfloor$ by $n/x$ introduces $-n$ (from the fractional parts summing to roughly $n/2$ each in the positive and negative direction)
• The Euler-Mascheroni constant $\gamma$ enters through $H_n = \ln n + \gamma + \ldots$

For JEE purposes: the key fact is $D(n) \sim n\ln n$, with a correction of order $\sqrt{n}$.

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Worked problems

Problem 1. Estimate $\displaystyle\sum_{k=1}^{1000} \tau(k)$.

$D(1000) \approx 1000 \ln 1000 + (2\gamma-1) \cdot 1000 \approx 1000 \times 6.908 + 0.1544 \times 1000 \approx 7062$.

Exact value: $D(1000) = 7069$. The approximation is off by 7 — error is $O(\sqrt{1000}) \approx 31$, well within bounds.

Problem 2. Prove that $\displaystyle\sum_{k=1}^{n} \tau(k) \geq n\ln n / 2$ for all $n \geq 1$.

Since $\tau(k) \geq 2$ for all $k \geq 2$ and $\tau(1)=1$... this is weaker than the asymptotic. Use the lattice point lower bound: for each $x \leq \sqrt{n}$, there are at least $\lfloor \sqrt{n} \rfloor$ values of $y$, giving $D(n) \geq m^2 \geq (\sqrt{n}-1)^2$... actually the asymptotic gives the result directly for large $n$.

Problem 3. Show that the average number of divisors of integers up to $n$ grows like $\ln n$.

Average $= D(n)/n \approx (n\ln n)/n = \ln n$. So a "typical" integer near $n$ has about $\ln n$ divisors — logarithmically many, not a fixed constant.

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The median of an $n \times n$ multiplication table is determined by where the hyperbola $xy = \text{median}$ cuts through exactly half the table entries. That geometric picture is in The Multiplication Table Median.