TITLE: Another integral for $\pi$
QUESTION [18 upvotes]: Here is a new integral for $\pi$. 

$$\int_{0}^{1}\sqrt{\frac{\left\{1/x\right\}}{1-\left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x} = \pi $$
  where $\left\{x\right\}$ denotes the fractional part of $x$.

Do you have any proof?

REPLY [3 votes]: This is just a (long) comment on Pranav Arora's fine answer.  I think there's a slightly easier way to get from
$$I=2\sum_{n=1}^\infty\int_0^\infty{x^2\over[(n+1)x^2+n][nx^2+(n-1)]}dx$$ 
to the telescoping sum.  
It's easy to check that
$$\begin{align}
{x^2\over[(n+1)x^2+n][nx^2+(n-1)]}&={n\over(n+1)x^2+n}-{n-1\over nx^2+(n-1)}\\
&={1\over\left({n+1\over n}\right)x^2+1}-{1\over\left({n\over n-1}\right)x^2+1}
\end{align}$$
From this and  the general arctangent integral
$$\int_0^\infty{1\over ax^2+1}dx={\pi\over2\sqrt a}$$
one can see directly that
$$I=2\sum_{n=1}^\infty\int_0^\infty{x^2\over[(n+1)x^2+n][nx^2+(n-1)]}dx=\pi\sum_{n=1}^\infty\left(\sqrt{n\over n+1}-\sqrt{n-1\over n} \right)$$