Calculate $\int_0^1{x·\lceil1/x\rceil dx}$ I am trying to calculate following integral:
$$\int_0^1{x·\biggl\lceil \frac{1}{x}\biggr\rceil dx}$$
I tried usual change t=1/x but not able to further advance.
Thanks! 
 A: $$\int_{1}^{\infty}\frac{\lceil x\rceil}{x^3}\,dx =\sum_{n\geq 1}\int_{n}^{n+1}\frac{n+1}{x^3}\,dx=\sum_{n\geq 1}\frac{2n+1}{2n^2(n+1)}=\color{red}{\frac{1}{2}+\frac{\pi^2}{12}}$$
holds by partial fraction decomposition.
A: If $\frac 1{n} \le x < \frac 1{n-1}$ then $\lceil \frac 1x \rceil = n$. and $\int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx= \int_{\frac 1{n}}^{\frac 1{n-1}} xn dx = n\frac {x^2}2|_{\frac 1n}^{\frac 1{n-1}}= \frac n2(\frac 1{(n-1})^2 -\frac 1{n^2}=\frac n2(\frac {2n-1}{n^2(n-1)^2})=\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}$
So $\int_{0}^{1} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty \int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty[\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}]=\sum\limits_{n=1}^\infty[\frac 1{n^2} - \frac {1}{2(n+1)n^2}]$
A: $$
\begin{aligned}
\int_0^1x\cdot\left\lceil \frac{1}{x}\right\rceil\; dx
&=
\sum_{k\ge 1}
\int_{1/(k+1)}^{1/k}x\cdot
\underbrace{\left\lceil \frac{1}{x}\right\rceil}_{\in (k,k+1)\text{ a.e}} \;dx
\\
&=
\sum_{k\ge 1}
\int_{1/(k+1)}^{1/k}x\cdot
(k+1)\;dx
\\
&=
\sum_{k\ge 1}
(k+1)
\left[\ \frac 12 x^2\ \right]_{1/(k+1)}^{1/k}
\\
&=
\frac 12
\sum_{k\ge 1}
(k+1)
\left[\  \frac 1{k^2}-\frac 1{(k+1)^2}\ \right]_{1/(k+1)}^{1/k}
\\
&=
\frac 12
\sum_{k\ge 1}
\frac {2k+1}{k^2(k+1)}
\\
&=
\frac 12
\sum_{k\ge 1}
\frac k{k^2(k+1)}
+
\frac 12
\sum_{k\ge 1}
\frac {k+1}{k^2(k+1)}
\\
&=
\frac 12
\sum_{k\ge 1}
\left[
\frac 1k-\frac 1{k+1}\right]
+
\frac 12
\sum_{k\ge 1}
\frac 1{k^2}
\\
&=
\frac 12\cdot 1+\frac 12\cdot\frac{\pi^2}6\ .
\end{aligned}
$$
Computer check, here PARI/GP:
? intnum( x=1.e-12, 1, x*ceil(1/x), 8 )
%12 = 1.3224146303918600341384604269228003273
? 1/2 + Pi^2/12
%13 = 1.3224670334241132182362075833230125946

