Another integral for $\pi$

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, $$\boxed{I=\pi}$$

Let me suggest a general approach.

We start with a theorem.

Theorem. (O. Oloa) Let $f$ be a piecewise continuous function on $(0,1)$ verifying $ \displaystyle \int_{0}^{1} \frac{ \left|f (x) \right| }{x} \: \mathrm{d}x <\infty.$ Then $$ \int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} = \int_{0}^{1} f(x) \frac{ \mathrm{d}x}{x} \qquad (*) $$ where $\left\{x\right\}$ denotes the fractional part of $x$.

Proof. One has \begin{align*} \int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} &= \sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left(\left\{ u \right\}\right) \: \frac{\mathrm{d} u}{u(u-1)} \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left(u-k\right) \: \frac{\mathrm{d} u}{u(u-1)} \\ &= \sum_{k=1}^{\infty} \int_{0}^{1} f \left(v\right) \: \frac{\mathrm{d} v}{(v+k)(v+k-1)} \\ &= \int_{0}^{1} f \left(v\right) \sum_{k=1}^{\infty} \frac{1}{(v+k)(v+k-1)} \mathrm{d} v \\ &= \int_{0}^{1} f(v) \frac{ \mathrm{d}v}{v},\end{align*} where the permutation between the infinite sum and the integration is allowed by the dominated convergence theorem: let $v \in (0,1)$, as $N$ tends to $\infty$, \begin{align} \displaystyle f \left(v\right) \sum_{k=1}^{N} \frac{1}{(v+k)(v+k-1)} & = f(v) \sum_{k=1}^{N}\left(\frac{1}{v+k-1}-\frac{1}{v+k}\right) = \frac{f(v)}{v}-\frac{f(v)}{v+N} \longrightarrow \frac{f(v)}{v} \nonumber \end{align} and one has \begin{align} \displaystyle \left|f \left(v\right) \sum_{k=1}^{N} \frac{1}{(v+k)(v+k-1)}\right| = \left| \frac{f(v)}{v}-\frac{f(v)}{v+N}\right| = \frac{\left| f(v) \right|}{v}-\frac{\left| f(v) \right|}{v+N} \leq \frac{ \left|f (v) \right| }{v} \nonumber \end{align} the latter function being such that $ \displaystyle \int_{0}^{1} \frac{ \left|f (v) \right| }{v} \: \mathrm{d}v <\infty$ by hypothesis. $\square$

Now, to get our result one may apply $(*)$ with $\displaystyle f(x)=\sqrt{\frac{x}{1-x}}$. Clearly $f$ satisfies the hypotheses of our Theorem and $$ \displaystyle \int_{0}^{1} \left| f(x) \right| \: \frac{ \mathrm{d}x}{x} = \int_{0}^{1} \left( \frac{x}{1-x}\right)^{1/2} \frac{ \mathrm{d}x}{x} = 2 \int_{0}^{1} \frac{1}{\sqrt{1-u^2}} \: \mathrm{d}u = \pi. $$

Similarly, inserting successively \begin{align} \displaystyle f(x):= \sqrt{\frac{x}{1-a^2 x}} \, ,\qquad 0 < a \leq 1 \nonumber \end{align} and
\begin{align}\displaystyle f(x):=\sqrt[n]{\frac{x}{1-x}} \, ,\qquad n=2,3, \cdots\nonumber \end{align} into $(*)$ gives some generalizations.

Proposition 1. Let $ 0 < a \leq 1$. Then \begin{align} \displaystyle \int_{0}^{1} \sqrt{\frac{\left\{1/x\right\}}{1-a^2 \left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x} & = \frac{2\arcsin a}{a} \end{align} where $\left\{x\right\}$ denotes the fractional part of $x$.

The initial integral is obtained with $a=1$.

Proposition 2. Let $n=2,3, \cdots .$ Then \begin{align} \displaystyle \int_{0}^{1} \sqrt[n]{\frac{\left\{1/x\right\}}{1-\left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x} & = \frac{\pi}{\sin(\pi/n)} \end{align} where $\left\{x\right\}$ denotes the fractional part of $x$.

The initial integral is obtained with $n=2$.

The (new) identity $(*)$ may be seen as a dual of the Gauss famous invariance result in ergodic theory:

Theorem. Let $f \in L^1(0,1)$. Then $$ \int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{x+1} = \int_{0}^{1} f(x) \frac{ \mathrm{d}x}{x+1} $$ where $\left\{x\right\}$ denotes the fractional part of $x$.

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

$$I=2\sum_{n=1}^\infty\int_0^\infty{x^2\over[(n+1)x^2+n][nx^2+(n-1)]}dx=\pi\sum_{n=1}^\infty\left(\sqrt{n\over n+1}-\sqrt{n-1\over n} \right)$$