Problem
Evalaute
\[\int_1^\infty\frac{\left\{x\right\}-\frac{1}{2}}{x}\,\mathrm dx\]where the curly braces denote the fractional part of $x$.
Solution
First we may rewrite the integral as a series,
\[\begin{align*}\int_1^\infty\frac{\left\{x\right\}-\frac{1}{2}}{x}\,\mathrm dx &=\sum_{n=1}^\infty\int_n^{n+1}\frac{\left\{x\right\}-\frac{1}{2}}{x}\,\mathrm dx \\ &=\sum_{n=1}^\infty\int_0^1\frac{\left\{x+n\right\}-\frac{1}{2}}{x+n}\,\mathrm dx \\ &=\sum_{n=1}^\infty\int_0^1\frac{x-\frac{1}{2}}{x+n}\,\mathrm dx \\ &= \sum_{n=1}^\infty\int_0^1\left(1-\left(n+\frac{1}{2}\right)\frac{1}{x+n}\right)\,\mathrm dx \\ &=\sum_{n=1}^\infty\left(1+\left(n+\frac{1}{2}\right)\ln(n)-\left(n+\frac{1}{2}\right)\ln(n+1) \right ) \\ &= \ln\left(\prod_{n=1}^\infty\left(\frac{n}{n+1} \right )^{n+1/2}e \right ).\end{align*}\]Examining the partial products yields the telescoping formula
\[\prod_{n=1}^m\left(\frac{n}{n+1} \right )^{n+1/2}e=\frac{m!e^m}{(m+1)^{m+1/2}}.\]If we apply Stirling’s approximation and take a limit we find that
\[\begin{align*}\lim_{m\to\infty}\prod_{n=1}^m\left(\frac{n}{n+1} \right )^{n+1/2}e &= \lim_{m\to\infty}\frac{m!e^m}{(m+1)^{m+1/2}} \\ &= \lim_{m\to\infty}\frac{\sqrt{2\pi m}\cdot m^m e^m}{e^m(m+1)^{m+1/2}} \\ &= \sqrt{2\pi}\lim_{m\to\infty}\sqrt{\frac{m}{m+1}}\left(\frac{m}{m+1} \right )^m\\ &= \sqrt{2\pi}\lim_{m\to\infty}\left(1-\frac{1}{m+1} \right )^m\\ &= \sqrt{2\pi}\lim_{m\to\infty}\left(1-\frac{1}{m+1} \right )^{m+1}\left(1-\frac{1}{m+1} \right )^{-1} \\ &= \sqrt{2\pi}\lim_{m\to\infty}\left(1-\frac{1}{m+1} \right )^{m+1} \\ &= \frac{\sqrt{2\pi}}{e}.\end{align*}\]So
\[\int_1^\infty\frac{\left\{x\right\}-\frac{1}{2}}{x}\,\mathrm dx=\frac{1}{2}\ln(2\pi)-1.\]