An interesting integral about γ

Question

It is well known that

$$\gamma = \lim_{n \to +\infty} (H_n − \ln(n)) = 0.5772(...)$$

where $H_n$ is the sum of reciprocals of all integers from $1$ to $n$.

Prove that $$\int_1^\infty \frac{\{ x\}}{x^2} dx = 1 - \gamma$$

Answer 1

Consider $\{x\} = x - \lfloor x \rfloor$, then:

\begin{equation*} \begin{split} \int_1^{+\infty} \dfrac{\{x\}}{x^2} \ \text{d}x & = \int_1^{+\infty} \left(x - \lfloor x \rfloor \right) \dfrac{\text{d}x}{x^2} \\\\ & = \lim_{n\to +\infty} \int_1^n \left(x - \lfloor x \rfloor\right) \dfrac{\text{d}x}{x^2} \\\\ & = \lim_{n\to +\infty} \ln(n) - \sum_{k = 1}^{n-1} \int_k^{k+1} \dfrac{k}{x^2}\ \text{d}x \\\\ & = \lim_{n\to +\infty} \ln(n) - \sum_{k = 1}^{n-1} k \left(\dfrac{1}{k} - \dfrac{1}{k+1}\right) \\\\ & = \lim_{n\to +\infty} \ln(n) - \sum_{k = 1}^{n-1} \dfrac{1}{k+1} \\\\ & = \lim_{n\to +\infty} \ln(n) - H_n + 1 \\\\ & = 1 - \gamma \end{split} \end{equation*}

Where $H_n$ is the $n-$th Harmonic number.