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 $x$.

  • 7
    $\begingroup$ Have you tried any methods of evaluating it? Perhaps in series form? $\endgroup$
    – B. Mehta
    Jul 19, 2018 at 16:31

3 Answers 3


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.

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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


  • $\begingroup$ This is the same solution as that of Marty Cohen, but explained my way. $\endgroup$
    – user65203
    Jul 19, 2018 at 21:00

You want $\int_{0}^{1}\sqrt{\bigg\{\frac{1}{x}\bigg\}}dx $.


$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 $g^{(-1)}(x)$.


$$\int_{0}^{1}\sqrt{\bigg\{\frac{1}{x}\bigg\}}dx=\\\sum _{n=1}^{\infty } \int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{\bigg\{\frac{1}{x}\bigg\}} \, dx=\\\sum _{n=1}^{\infty } \int_{\frac{1}{n+1}}^{\frac{1}{n}} \sqrt{\frac{1}{x}-n} \, dx=\\\sum _{n=1}^{\infty } \frac{-4 \sqrt{n}+\pi +n \pi -2 (1+n) \sin ^{-1}\left(\frac{-1+n}{1+n}\right)}{4 \sqrt{n} (1+n)}$$

With the aid of computer algebra, we obtain integral and approximate series with value.

Probably closed form of the sum does not exist.


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


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