Evaluate $$\int_1^{\infty} \frac {(\{x\}-\frac 12)\,dx}{x}$$ where $\{x\}$ represents fractional part of $x$.
My attempt :
$$\int_1^{\infty} \frac {(\{x\}-\frac 12)\,dx}{x}=\int_1^{\infty} \frac {(x-\lfloor x\rfloor-\frac 12)\,dx}{x}$$ $$=\lim_{n\to \infty} \left(\int_1^n \,dx-\int_1^n \frac {(\lfloor x\rfloor +\frac 12)\,dx}{x}\right)$$
$$=\lim_{n\to \infty} \left((n-1)-\frac 12\left[\ln\left(\frac {2^3}{1^3}\cdot\frac {3^5}{2^5}\cdot\frac {4^7}{3^7}\cdots \frac {n^{2n-1}}{(n-1)^{2n-1}}\right) \right]\right)$$
$$=-1+\lim_{n\to \infty} n-\frac 12 \ln\left (\frac {n^{2n+1}}{(n!)^2}\right) $$
I am stuck up here and can't resolve further. Any help would be greatly appreciated
Gotcha I got it.
$$L=-1+\lim_{n\to \infty} n-\frac 12 \ln\left (\frac {n^{2n+1}}{(n!)^2}\right) $$
Using Stirling's approximation of factorials $$L=-1+\lim_{n\to \infty} n-\ln\left(\frac {n^{n+\frac 12}}{\sqrt {2\pi}\cdot \frac {n^{n+\frac 12}}{e^n}}\right) =-1+\ln(\sqrt {2\pi})$$