How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-k\right) dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\ln(1+\frac{1}{k})-\frac{1}{k+1}\right)\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)\\&=1.
\end{aligned}
\end{equation}
Please give me some ideas,thank you!
 A: With the help of reuns,I can complete the proof.
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-k\right) dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\ln(1+\frac{1}{k})-\frac{1}{k+1}\right)\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{2k^2}-\frac{1}{k+1}\right)\\&=\frac{1}{2}.
\end{aligned}
\end{equation}
Since
$$\frac{1}{n}=\sum_{k=n}^\infty \frac{1}{k(k+1)}<\sum_{k=n}^\infty \frac{1}{k^2}<\sum_{k=n}^\infty \frac{1}{k(k-1)}=\frac{1}{n-1}$$
and
$$\frac{1}{n^3}\leq\sum_{k=n}^\infty\frac{1}{k^3}\leq\frac{1}{n}\sum_{k=n}^\infty\frac{1}{k^2}<\frac{1}{n(n-1)}.$$
Thanks to Martin R for reminding.
The following part is by Adam Latosiński, I just modified the typographical errors.Thanks for his help!
We still need to show that $\lim\limits_{x\rightarrow 0} \frac{F(x)}{x} = \frac{1}{2}$ also when we approach by a sequence with $x\neq\frac{1}{n}$. Let $n=[ 1/x]$, so that $\frac{1}{n+1}<x\le\frac{1}{n}$, that is $n\le\frac{1}{x}<{n+1}$. We have
\begin{align} \left|\frac{F(1/n)}{1/n} - \frac{F(x)}{x}\right| &\le \left|\frac{F(1/n)}{1/n} - \frac{F(1/n)}{x}\right| + \left|\frac{F(1/n)}{x} - \frac{F(x)}{x}\right| \\
&\le F(1/n) \left|n-\frac{1}{x}\right| + (n+1) |F(1/n)-F(x)|  \\
&\le F(1/n) + (n+1) \int_{x}^{\frac{1}{n}} \Big(\frac{1}{t}- \left[\frac{1}{t}\right]\Big) dt  \\
&\le F(1/n) + (n+1) \int_{x}^{\frac{1}{n}} 1 \,dt  \\
&\le F(1/n) + (n+1)(\frac{1}{n}-\frac{1}{n+1}) \rightarrow 0\end{align}
which proves that $$ \lim_{x\rightarrow 0} \frac{F(x)}{x} = \lim_{n\rightarrow\infty} nF(\frac{1}{n}) = \frac12$$
A: By the Stolz-Cesaro theorem
\begin{eqnarray} \lim_{n\to\infty}n F\left(\frac{1}{n}\right) &=&\lim_{n\to\infty}\frac{F\left(\frac1n\right)}{\frac1n}\\
&=&\lim_{n\to\infty}\frac{F\left(\frac1{n+1}\right)-F\left(\frac1n\right)}{\frac1{n+1}-\frac1n}\\ 
&=&-\lim_{n\to\infty}n(n+1)\int^{\frac{1}{n+1}}_{\frac{1}{n}} \left(\frac{1}{t} - n\right)dt \\
&=&-\lim_{n\to\infty}n(n+1)\int^{\frac{1}{n+1}}_{\frac{1}{n}} \left(\frac{1}{t} - n\right)dt \\
&=&-\lim_{n\to\infty}n(n+1)\left[\ln\left(\frac{n}{n+1}\right)-\frac1{n+1}\right] \\
&=&\frac12.
\end{eqnarray}
A: Using $\{x\}=x-\lfloor x\rfloor$ and substituting $t\mapsto\frac1t$, then integrating by parts (multiple times for higher precision), we get
$$
\begin{align}
\int_0^x\left(\frac1t-\left\lfloor\frac1t\right\rfloor\right)\mathrm{d}t
&=\int_{1/x}^\infty\left(\frac1{2t^2}+\frac{\{t\}-\frac12}{t^2}\right)\mathrm{d}t\\
&=\frac x2+\int_{1/x}^\infty\frac1{t^2}\,\mathrm{d}\left(\tfrac12\{t\}^2-\tfrac12\{t\}+\tfrac1{12}\right)\\
&=\frac x2-x^2\left(\tfrac12\left\{\tfrac1x\right\}^2-\tfrac12\left\{\tfrac1x\right\}+\tfrac1{12}\right)+\int_{1/x}^\infty\frac{\{t\}^2-\{t\}+\tfrac16}{t^3}\,\mathrm{d}t\\[6pt]
&=\frac x2-x^2\left(\tfrac12\left\{\tfrac1x\right\}^2-\tfrac12\left\{\tfrac1x\right\}+\tfrac1{12}\right)+O\!\left(x^3\right)
\end{align}
$$
Therefore,
$$
\lim_{x\to0}\frac1x\int_0^x\left(\frac1t-\left\lfloor\frac1t\right\rfloor\right)\mathrm{d}t=\frac12
$$
A: I have a solution that utilises the digamma function:
\begin{align}  F(\frac{1}{n}) &= \sum_{k=n}^\infty \int_{\frac{1}{k+1}}^{\frac{1}{k}} (\frac{1}{t} - k)dt = \\
&= \sum_{k=n}^\infty \int_{k}^{k+1} \frac{s - k}{s^2} ds = \\
&= \sum_{k=n}^\infty \int_{0}^{1} \frac{s}{(s+k)^2} ds = \\
&= \sum_{k=0}^\infty\int_{0}^{1} s \frac{1}{(s+k+n)^2} ds = \\
&= \int_{0}^{1} s \,\psi'(s+n) ds = \\
&= [s\psi(s+n)]_{s=0}^1 - \int_{0}^{1} \psi(s+n) ds = \\
&= \psi(n+1) - [\ln \Gamma(s+n)]_{s=0}^1 = \\
&= \psi(n+1) - \ln \frac{\Gamma(n+1)}{\Gamma (n)} = \\
&= \psi(n+1) - \ln n\end{align}
For large $n$ we have Stirling formula  $\ln\Gamma(n+1) = (n+\frac12)\ln n - n + \mathcal O(n^{-1})$, so $\psi (n+1) = \ln n + \frac{1}{2n} + \mathcal O(n^{-2}) $, which gives
$$ \lim_{n\rightarrow\infty} nF(\frac{1}{n}) = \frac12$$
We still need to show that $\lim_{x\rightarrow 0} \frac{F(x)}{x} = \frac{1}{2}$ also when we approach by a sequence with $x\neq\frac{1}{n}$. Let $n=\lfloor 1/x\rfloor$, so that $\frac{1}{n+1}<x\le\frac{1}{n}$, that is $n\le\frac{1}{x}<\frac{1}{n+1}$. We have
\begin{align} \left|\frac{F(1/n)}{1/n} - \frac{F(x)}{x}\right| &\le \left|\frac{F(1/n)}{1/n} - \frac{F(1/n)}{x}\right| + \left|\frac{F(1/n)}{x} - \frac{F(x)}{x}\right| \le \\
&\le F(1/n) \left|n-\frac{1}{x}\right| + \frac{1}{n} |F(1/n)-F(x)| \le \\
&\le F(1/n) + \frac{1}{n} \int_{x}^{\frac{1}{n}} \Big(\frac{1}{t}- \lfloor\frac{1}{t}\rfloor\Big) dt \le \\
&\le F(1/n) + \frac{1}{n} \int_{x}^{\frac{1}{n}} 1 \,dt = \\
&\le F(1/n) + \frac{1}{n}(\frac{1}{n}-\frac{1}{n+1}) \rightarrow 0\end{align}
which proves that $$ \lim_{x\rightarrow 0} \frac{F(x)}{x} = \lim_{n\rightarrow\infty} nF(\frac{1}{n}) = \frac12$$
A: i) Consider $$ A_n :=\int^{\frac{1}{n-1}}_{ \frac{1}{n} } \
\frac{1}{x}\ dx - (n-1) \bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg] $$
Hence $F(\frac{1}{n-1})=A_n +A_{n+1}+\cdots $
So $$ A_n\bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg]^{-1} \rightarrow
\frac{1}{2}$$
so that $\frac{F(\frac{1}{n-1}) }{ \frac{1}{n-1} } \rightarrow
\frac{1}{2} $
ii) $ \frac{F(\frac{1}{n} )}{\frac{1}{n-1}}\leq \frac{F(x)}{x} \leq
\frac{F(\frac{1}{n-1} )}{\frac{1}{n}} $ where $\frac{1}{n}\leq
x<\frac{1}{n-1} $ 
