It can be shown that for every integer $p\geq 0$, this integral identity holds:
\begin{align}
\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx &= -1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)\\
&= -1+\frac{\zeta(2)+\zeta(3)+\zeta(4)+\cdots+\zeta(p+2)}{p+1}
\end{align}
Pretty neat! I find it interesting that the quantity
$$\frac{\zeta(2)+\zeta(3)+\zeta(4)+\cdots+\zeta(p+2)}{p+1}$$
is precisely the arithmetic mean of the numbers $\zeta(2),\zeta(3),\zeta(4),\dots,\zeta(p+2)$. Anyway, notice that the integral
$$\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx$$
also converges for $p=-1$. This is intuitively plausible because $\lfloor x\rfloor$ grows roughly like $x$, so
$$\frac{1}{x^{-1+2}\lfloor x\rfloor}=\frac{1}{x\lfloor x\rfloor}$$
behaves like $1/x^2$ as $x\to\infty$, yielding a convergent integral. To clear up any doubts, I've left a proof at the bottom of the post.
Now, the equation
$$\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx=-1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)$$
doesn't make sense for $p=-1$ because substituting $-1$ for $p$ involves division by zero. Nevertheless, given the obvious connection between
$$\int_1^\infty\frac{1}{x^{p+2}\lfloor x\rfloor}\text{ }dx$$
and $\zeta(n)$ for $n\in\mathbb{N}$, I can't help but wonder if there's a link between
\begin{align}
\int_1^\infty\frac{1}{x\lfloor x\rfloor}\text{ }dx &= \lim_{k\to\infty}\left(\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\right)\\
&= \lim_{k\to\infty}\left(\sum_{n=1}^k \frac{\ln(n+1)-\ln(n)}{n}\right)\\
&= \sum_{n=1}^\infty \frac{\ln(n+1)-\ln(n)}{n}
\end{align}
and the zeta function. Maybe it’s the exact value of zeta evaluated at some other point in the complex plane?
Seeing the pattern in the identity, my only "reasonable" idea was to try and find a link between the sum the series (I'll call it $S$) and $\zeta(1)$. This expression is undefined ($\zeta$ has a singularity at $1$), but its principal value exists and equals the Euler-Mascheroni constant:
$$\lim_{h\to 0}\frac{\zeta(1-h)+\zeta(1+h)}{2}=\gamma$$
We can't have an exact equality; $\gamma\approx 0.57$, but WolframAlpha says that $S\approx 1.25$. It's unlikely, but could these numbers be related in some other way? If not, are there any other possible links between $S$ and the zeta function? Any answer is greatly appreciated.
Proof: the sequence
$$\left\lbrace\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\right\rbrace_{k=0}^\infty$$
is strictly increasing because the integrand is strictly positive over $[1,k+1]$. Now,
\begin{align}
\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx &= \int_1^2\frac{1}{x\lfloor x\rfloor}\text{ }dx+\int_2^3\frac{1}{x\lfloor x\rfloor}\text{ }dx+\cdots+\int_k^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\\
&= \sum_{n=1}^k \int_n^{n+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\\
&= \sum_{n=1}^k \int_n^{n+1}\frac{1}{x\cdot n}\text{ }dx\\
&= \sum_{n=1}^k \frac{1}{n}\int_n^{n+1}\frac{1}{x}\text{ }dx\\
&= \color{green}{\sum_{n=1}^k \frac{\ln(n+1)-\ln(n)}{n}}\\
&= \sum_{n=1}^k \frac{1}{n}\ln\left(\frac{n+1}{n}\right)\\
&= \sum_{n=1}^k \frac{1}{n}\ln\left(1+\frac{1}{n}\right)\\
&< \sum_{n=1}^k \frac{1}{n}\cdot\frac{1}{n}\\
&< \sum_{n=1}^\infty \frac{1}{n^2}
\end{align}
so the terms of the sequence are bounded above. It follows that
$$\left\lbrace\int_1^{k+1}\frac{1}{x\lfloor x\rfloor}\text{ }dx\right\rbrace_{k=0}^\infty$$
is a convergent sequence. $\blacksquare$
Best Answer
If we expand the logarithm into its Maclaurin series and change the order of summations, we obtain $$ \sum\limits_{k = 1}^\infty {\frac{1}{n}\log \left( {1 + \frac{1}{n}} \right)} = \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} }}{k}\zeta (k + 1)} = \int_0^1\frac{\psi (z + 1) + \gamma}{z}dz $$ where $\psi$ is the digamma function. The integral shares some similarities with $(25.5.17)$. A faster converging version is $$ \log 2 + \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} }}{k}(\zeta (k + 1) - 1)} $$ giving the value $1.257746887\ldots$. See also Integral of digamma function.