Integral Expression of Harmonic Number
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Theorem
Let $\sequence {H_n}_{n \mathop \in \N}$ be the harmonic numbers.
Then:
- $\ds H_n = 1 + \int_1 ^n \dfrac {\floor u} {u^2} \rd u$
where $\floor u$ denotes the floor of $u$.
Proof
Observe that:
- $\ds \forall x \in \R_{\ge 1} : \floor x = \sum_{1 \mathop \le k \mathop \le x} 1$
Let $f: \R_{\ge 1} \to \R$ be defined as:
- $\ds \map f x := \dfrac 1 x$
Then:
\(\ds \sum_{k \mathop = 1}^n \frac 1 k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n 1 \cdot \map f k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \floor n \map f n - \int_1^n \floor u \map {f'} u \rd u\) | Abel's Summation Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \int_1 ^n \dfrac {\floor u} {u^2} \rd u\) | Derivative of Reciprocal |
$\blacksquare$