Another integral for $\pi$ 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?

With the substitution $1/x=t$, the integral is:
$$I=\int_1^{\infty} \sqrt{\frac{\{t\}}{1-\{t\}}}\frac{dt}{t(t-1)}=\sum_{n=1}^{\infty}\int_n^{n+1}\sqrt{\frac{t-n}{n+1-t}}\frac{dt}{t(t-1)}$$
Next, use the substitution $\dfrac{t-n}{n+1-t}=x^2 \Rightarrow t=n+1-\dfrac{1}{1+x^2} \Rightarrow dt=\dfrac{2x}{(1+x^2)^2}\,dx$ i.e
$$I=2\sum_{n=1}^{\infty} \int_0^{\infty} \frac{x^2}{(x^2(n+1)+n)(nx^2+n-1)}\,dx$$
$$\Rightarrow I=2\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\int_0^{\infty}\frac{x^2}{\left(x^2+\frac{n}{n+1}\right)\left(x^2+\frac{n-1}{n}\right)}\,dx$$
The above integral can be evaluated by decomposing into partial fractions i.e
$$\begin{aligned}
I &=\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\frac{\pi}{\left(\sqrt{\frac{n}{n+1}}+\sqrt{\frac{n-1}{n}}\right)} \\
&=\pi\sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}(n+\sqrt{n^2-1})}\\
&=\pi\sum_{n=1}^{\infty} \frac{n-\sqrt{n^2-1}}{\sqrt{n(n+1)}} \\
&=\pi \sum_{n=1}^{\infty} \left(\sqrt{\frac{n}{n+1}}-\sqrt{\frac{n-1}{n}}\right)
\end{aligned}$$
The final sum telescopes and its value is $1$. Hence,
$$\boxed{I=\pi}$$
