Evaluate $\int_1^{\infty} \frac {(\{x\}-\frac 12)dx}{x}$

Question

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

@openspace That is what I did for the second Integral and hence got the log term — Rohan Shinde, Jun 01 '18 at 17:45
See this post from 2014: https://math.stackexchange.com/questions/1054064/is-there-other-methods-to-evaluate-int-1-infty-frac-x-frac12x?rq=1
It should contain all the help you need. — Umberto P., Jun 01 '18 at 17:45
@Manthanein as I see , you used $[1,n]$ instead of $[n,n+1]$ — openspace, Jun 01 '18 at 18:11

score 4 · Answer 1 · answered Jun 01 '18 at 17:59

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 $$L=-1+\lim_{n\to \infty} n-\ln\left(\frac {n^{n+\frac 12}}{\sqrt {2\pi}\cdot \frac {n^{n+\frac 12}}{e^n}}\right) =-1+\ln(\sqrt {2\pi})$$

Evaluate $\int_1^{\infty} \frac {(\{x\}-\frac 12)dx}{x}$

1 Answers1