Who's up to the challenge? 12

Rewrite the integral as:

$\displaystyle \int_1^{\infty} \frac{\{x\}-\frac{1}{2}}{x} dx = \int_1^{\infty} \frac{x-\lfloor x\rfloor -\frac{1}{2}}{x} dx$

$\displaystyle =\lim_{k\to \infty} \sum_{m=1}^{k} \int_m^{m+1} \frac{x - m -\frac{1}{2}}{x} dx$

Now it's just basic integration:

$\displaystyle = \lim_{k\to \infty} \sum_{m=1}^{k} \bigg(1+(m+\frac{1}{2} ) \ln (m) - (m+\frac{1}{2} ) \ln (m+1) \bigg)$

$\displaystyle = \lim_{k\to \infty} \sum_{m=1}^{k} \bigg(1+(m-1+1+\frac{1}{2} ) \ln (m) - (m+\frac{1}{2} ) \ln (m+1) \bigg)$

$\displaystyle = \lim_{k\to \infty} \sum_{m=1}^{k} \bigg(1+\ln (m) + \underbrace{(m-1+\frac{1}{2} ) \ln m - (m+\frac{1}{2} ) \ln (m+1)}_\text{Telescoping} \bigg)$

$\displaystyle =\lim_{k\to \infty} \big( k+\ln (k!) -(k+\frac{1}{2} ) \ln (k+1) \big)$

$\displaystyle =\lim_{k\to \infty} \Bigg( \ln (e^k) +\ln (k!) - \ln \bigg((k+1)^{(k+\frac{1}{2} )}\bigg) \Bigg)$

Write as a single Logarithm:

$\displaystyle = \lim_{k\to \infty} \ln \Bigg(\frac{k! e^k}{(k+1)^{(k+\frac{1}{2} )} } \Bigg)$

Use Stirling's formula for very large arguments of the factorial:

$\displaystyle = \lim_{k\to \infty} \ln \Bigg(\frac{\sqrt{2\pi k} (\frac{k}{e})^k e^k}{(k+1)^{(k+\frac{1}{2} )} } \Bigg)$

$\displaystyle = \lim_{k\to \infty} \ln \Bigg(\frac{\sqrt{2\pi} k^{k+\frac{1}{2}} }{(k+1)^{(k+\frac{1}{2} )} } \Bigg)$

$\displaystyle = \lim_{k\to \infty} \ln \Bigg(\frac{\sqrt{2\pi}}{ (1+\frac{1}{k})^{k+\frac{1}{2}}} \Bigg)$

By the definition of the Euler Number $e$ :

$\displaystyle = \ln (\frac{\sqrt{2\pi}}{e}) = \boxed{\ln (\sqrt{2\pi} ) - 1}$