Evaluate the following integral $ \int_1^{\infty} \frac{\lbrace x\rbrace-\frac{1}2}{x} dx$ $$\int_1^{\infty} \frac{\lbrace x\rbrace-\frac{1}2}{x} dx$$
Here $\lbrace\cdot\rbrace$ denotes the fractional part.
I found this challenging integral, and I'm curious about the solution, so I decided to do some efforts to solve it, but sadly I didn't, any hints?
Attempts:
\begin{align}
\int_1^{\infty} \frac{\lbrace x\rbrace-\frac{1}2}{x} dx&=\int_1^{\infty} \frac{1\lbrace x\rbrace-1}{2x} dx\\
&=\int_1^{\infty}\frac{\lbrace x\rbrace}{x}-\frac{1}{2x}dx\\
&=\int_1^\infty \frac{x-\lfloor x\rfloor-1}{x}-\frac{1}{2x}dx\\
&=\int_1^\infty \frac{x-\lfloor x\rfloor-1}{x} dx -\int_1^\infty \frac{dx}{2x}
\end{align}
I thought about this property:
$$\int_0^\infty \varphi (x) dx=\lim_{a\to \infty} \int_0^a \varphi(x) dx$$
So I applied it only for the second fraction because its antiderivative was easy enough, and here's what I've got:
\begin{align}
\int_1^\infty \frac{dx}{2x}&=\lim_{a\to \infty} \int_1^a \frac{dx}{2x}\\
&=\lim_{a\to \infty}\frac{\ln (x)}{2}\bigg\vert_0^a\\
&=\lim_{a\to \infty}\frac{\ln (a)}2 -\frac{\ln (0)}{2}
\end{align}
And here I felt that I'm wrong I can't get $\infty -\infty$, So any thoughts or hints, I'll be thankfull!
 A: Not to take away from @Whatsup's clever answer, but I did it another way.
Start with the integral formula, valid for $\Re(s)>0$:
$$
\zeta(s) = \frac{s}{s-1} - s\int _1^{\infty}\frac{\{x\}}{x^{s+1}}\,dx
$$Introduce the $1/2$ in the integrand:
$$
\zeta(s) = \frac{s}{s-1} - s\int _1^{\infty}\frac{\{x\}-1/2+1/2}{x^{s+1}}\,dx
$$$$
\zeta(s) = \frac{s}{s-1} - s\int _1^{\infty}\frac{\{x\}-1/2}{x^{s+1}}\,dx - \frac{1}{2}
$$Solve for the integral:
$$
\int _1^{\infty}\frac{\{x\}-1/2}{x^{s+1}}\,dx = \frac{-2 (s-1) \zeta (s)+s+1}{2 (s-1) s}
$$Take the limit as $s\to 0^+$; the LHS exists by Dirichlet's Test and the RHS can be evaluated using L'Hôpital's Rule.
$$
\int _1^{\infty}\frac{\{x\}-1/2}{x}\,dx =\lim_{s\to 0^+} \frac{-2 (s-1) \zeta (s)+s+1}{2 (s-1) s}
$$
$$
=\lim_{s\to 0^+} \frac{-2 (s-1) \zeta '(s)-2 \zeta (s)+1}{4s-2}
$$Using $\zeta(0)=-1/2$ and $\zeta'(0)=-1/2\log(2 \pi)$ as shown here gives the same result.
$$
=\lim_{s\to 0^+} \frac{-2 (s-1) (-1/2\log(2 \pi))-2 (-1/2)+1}{4s-2}
$$
$$
=1/2\log(2 \pi)-1= \log(\sqrt{2\pi}/e)
$$
A: This function is not integrable in the Lebesgue sense, so you can only evaluate the Cauchy principal value.
That is, what you want to evaluate is the limit $$\lim_{M \rightarrow +\infty} \int_1^M \frac{\{x\}-\frac12}xdx.$$
It is easy to see that it suffices to take the limit for integer values of $M$. We first compute, for every positive integer $k$:
$$\int_k^{k + 1}\frac{\{x\}-\frac12}xdx = \int_k^{k + 1}\frac{x- k-\frac12}xdx = 1 - \left(k + \frac 1 2\right) (\ln(k + 1) - \ln k).$$
We then take the sum:
$$\int_1^{M + 1} \frac{\{x\}-\frac12}xdx = \sum_{k = 1}^M\left(1 - \left(k + \frac 1 2\right) (\ln(k + 1) - \ln k)\right).$$
This simplifies to: $$M - \left(M + \frac12\right)\ln(M + 1) + \ln M!$$ which, by Stirling's formula, converges to $\ln\frac{\sqrt{2\pi}}e\approx-0.0810614668$.
A: Euler-Maclaurin formula applied to $\log N!$ gives
$$
\begin{aligned}
\log N!
&=\int_1^N\log x\mathrm dx+{\log N\over2}+\int_1^N{\overline B_1(x)\over x}\mathrm dx \\
&=\left(N+\frac12\right)\log N-N+1+\int_1^N{\overline B_1(x)\over x}\mathrm dx
\end{aligned}
$$
However, according to Stirling's approximation, we have
$$
\log N!=\left(N+\frac12\right)\log N-N+\frac12\log2\pi+\mathcal O\left(\frac1N\right)
$$
Consequently as $N\to\infty$ we get
$$
\int_1^\infty{\overline B_1(x)\over x}\mathrm dx=\int_1^\infty{\{x\}-1/2\over x}\mathrm dx=-1+\frac12\log2\pi
$$
