Square root of fractional part integral Does the following integral have a closed form ? 
$$\int_{0}^{1}\sqrt{\bigg\{\frac{1}{x}\bigg\}}dx$$
Where $\{x\}$ denotes the fractional part of $x$.
 A: Due to the complexity of the computation (see other answers), we consider integrating $x\,dy$ instead of $y\,dx$. 
The function has discontinuities at every $x=\dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $\dfrac1{n+1}$ and $\dfrac1n$ using
$$y=\sqrt{\frac1x-n}\iff x=\frac1{y^2+n}.$$
From this, the area under a tooth is given by
$$I_n=\int_0^1\frac{dy}{y^2+n}-\frac1{n+1}=\frac1{\sqrt n}\arctan\frac1{\sqrt n}-\frac1{n+1}.$$
The total area is the sum of the $I_n$, which doesn't seem to have a closed form.

If we use the Taylor development of the arc tangent,
$$\frac1{\sqrt n}\left(\frac1{\sqrt n}-\frac1{3n\sqrt n}+\frac1{5n^2\sqrt n}-\cdots\right)-\frac1{n+1}.$$
Now if we sum on $n$, noticing that the first term will telescope with the last one, we get
$$1-\frac13\zeta(2)+\frac15\zeta(3)-\frac17\zeta(4)+\cdots$$
A: You want
$\int_{0}^{1}\sqrt{\bigg\{\frac{1}{x}\bigg\}}dx
$.
Consider
$I
=\int_{0}^{1}f(\{1/x\})dx
$
where
$f(0) = 0.
f(1) = 1,
f'(x) \ge 0
$.
Since
$n \le 1/x
\le n+1
$
for
$1/(n+1)
\le x 
\le 1/n
$,
$\begin{array}\\
I
&=\int_{0}^{1}f(\{1/x\})dx\\
&=\sum_{n=1}^{\infty}\int_{1/(n+1)}^{1/n}f(\{1/x\})dx\\
&=\sum_{n=1}^{\infty}\int_{1/(n+1)}^{1/n}f(1/x-n)dx\\
&=\sum_{n=1}^{\infty}\int_{n}^{n+1}f(y-n)dy/y^2
\qquad y=1/x, x=1/y, dx=-dy/y^2\\
&=\sum_{n=1}^{\infty}\int_{0}^{1}f(y)dy/(y+n)^2\\
&=\int_{0}^{1}f(y)dy\sum_{n=1}^{\infty}\dfrac1{(y+n)^2}\\
\end{array}
$
so your result is
$\begin{array}
I
&=\sum_{n=1}^{\infty}\int_{0}^{1}f(y)dy/(y+n)^2\\
&=\sum_{n=1}^{\infty}\int_{0}^{1}\sqrt{y}dy/(y+n)^2\\
&=\sum_{n=1}^{\infty}\left(\dfrac{\cot(\sqrt{n})}{\sqrt{n}} - \dfrac1{n + 1}\right)
\qquad\text{(according to Wolfy)}\\
\end{array}
$
Note:
You can do this
for any function
(instead of $1/x$)
such that
$g(1) = 1,
g'(x) > 0,
g(x) \to \infty$
as $x \to 0$.
You have to look
at $g^{(-1)}(x)$.
A: $$\int_{0}^{1}\sqrt{\bigg\{\frac{1}{x}\bigg\}}dx=\\\sum _{n=1}^{\infty } \int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{\bigg\{\frac{1}{x}\bigg\}}
   \, dx=\\\sum _{n=1}^{\infty } \int_{\frac{1}{n+1}}^{\frac{1}{n}} \sqrt{\frac{1}{x}-n} \, dx=\\\sum _{n=1}^{\infty } \frac{-4 \sqrt{n}+\pi +n \pi -2 (1+n) \sin
   ^{-1}\left(\frac{-1+n}{1+n}\right)}{4 \sqrt{n} (1+n)}$$
With the aid of computer algebra, we obtain integral and approximate series with value.
Probably closed form of the sum does not exist.

Supplement:
Borrowing sum from user: Yves Daoust we can write:
$$\sum _{n=1}^{\infty } \left(\frac{\cot ^{-1}\left(\sqrt{n}\right)}{\sqrt{n}}-\frac{1}{n+1}\right)=1-\gamma -\int_0^1 \frac{\psi (1+x)}{2 \sqrt{x}} \, dx\approx 0.60204991$$
where: $\psi \left(1+x\right)$ is PolyGamma function
