Computing $\int_0^{\large1/1^2} \left\{ \frac{1}{t} \right\}\,\mathrm{d}t+\int_0^{\large 1/2^2} \left\{ \frac{1}{t} \right\}\,\mathrm{d}t+\cdots $ What tools would you recommend me for computing this series?
$$\int_0^{\large1/1^2} \left\{ \frac{1}{t} \right\}\,\mathrm{d}t+\int_0^{\large 1/2^2} \left\{ \frac{1}{t} \right\}\,\mathrm{d}t+\int_0^{\large 1/3^2} \left\{ \frac{1}{t} \right\}\,\mathrm{d}t+\cdots $$
$\{x\}$ is the fractional part of $x$
 A: Letting $x=1/t$, you get: $$\int_{0}^{1/k^2} \left\{\frac1t\right\}dt = \int_{k^2}^\infty \frac{1}{x^2}\{x\}dx = \sum_{i=k^2}^\infty \int_{0}^1 \frac u{(u+i)^2} du$$
and:
$$\begin{align} \int_{0}^1 \frac u{(u+i)^2} du &=\int_{0}^1 \left(\frac{1}{u+i}-\frac{i}{(u+i)^2}\right)dx\\
&=\log(i+1)-\log(i) +\frac{i}{i+1}-\frac{i}{i}\\&=\log\left(1+\frac{1}{i}\right) -\frac{1}{i+1}
\end{align}$$
So $$\sum_{i=k^2}^\infty \left(\log\left(1+\frac{1}{i}\right)-\frac{1}{i+1}\right)$$
Note that: $$\sum_{i=1}^\infty \left(\log\left(1+\frac{1}{i}\right)-\frac{1}{i+1}\right)=-\gamma+1$$
so $$\sum_{i=k^2}^\infty \left(\log\left(1+\frac{1}{i}\right)-\frac{1}{i+1}\right) = -\gamma+1-\left(\log(k^2) -( H_{k^2}-1)\right) =-\gamma-\log(k^2)+H_{k^2}$$
Then $$\sum_{k=1}^\infty \left(H_{k^2}-\gamma-\log(k^2)\right)$$
is the integral you want.
I'm not sure how much better than that you can do.
From the first formula here:
$$\gamma = H_n -\log(n)-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$ we see that:
$$H_{k^2}-\gamma-\log(k^2) = \frac{1}{2k^2}+O\left(\frac{1}{k^4}\right)$$ so we know the series converges.
If we take:
$$\begin{align}\sum_{k=1}^N \left(H_{k^2}-\gamma-\log(k^2)\right) &=
\left(\sum_{k=1}^N\sum_{j=1}^{k^2}\frac{1}{j}\right) -N\gamma - 2\log(N!)\\
&=\sum_{j=1}^{N^2}\frac{N+1-\lceil\sqrt{j}\rceil}{j}-N\gamma -2\log(N!)\\
&= (N+1){H_{N^2}} -N\gamma -2\log(N!) - \sum_{j=1}^{N^2} \frac{\lceil\sqrt {j}\rceil}{j}
\end{align} $$
We have
 $$\begin{align}(N+1)H_{N^2}-N\gamma &= (N+1)\left(H_{N^2}-\gamma\right)+\gamma \\
&= \gamma + (N+1)\left(\log(N^2) +\frac{1}{2N^2}-\frac{1}{12N^4}+\dots\right)\\
&=\gamma +2(N+1)\log(N) +O\left(\frac 1N\right)
\end{align}
$$
And $$2\log(N!) = (2N+1)\log N - 2N + \log(2\pi) + O\left(\frac{1}N\right)$$
So $$(N+1){H_{N^2}} -N\gamma -2\log(N!) =\gamma+2N+\log N - \log(2\pi) +O\left(\frac{1}{N}\right)$$
If you can get a good estimate for $\sum_{j=1}^K \frac{\lceil\sqrt {j}\rceil}{j}$ (within $o(1)$) you might be able to compute the limit here, but that seems difficult.
Since the entire sum is convergent, we know $\sum_{j=1}^{N^2} \frac{\lceil\sqrt {j}\rceil}{j}=2N+\log N + O(1)$, but that $O(1)$ hides the final value.
