Trigonometric integral involving the fractional part function

Question

I have tried the $u$ substitution yet I could not move forwards on this problem. Could you calculate in closed-form this integral?

$$\int_{0}^{\pi/2}\bigg\{\csc(x)\bigg\}\mathrm{dx}$$

Answer 1

Enforcing the substitution $x\mapsto \arcsin(x)$ followed by $x\mapsto 1/x$ we see that

$$\begin{align} \int_0^{\pi/2} \bigg\{\frac{1}{\sin(x)}\bigg\}\,dx&=\int_0^1 \bigg\{\frac1x\bigg\}\frac1{\sqrt{1-x^2}}\,dx\\\\ &=\int_1^\infty \{x\} \frac{1}{x\sqrt{x^2-1}}\,dx\\\\ &=\int_1^\infty \frac{x-\lfloor x\rfloor}{x\sqrt{x^2-1}}\,dx\\\\ &=\sum_{k=1}^\infty \int_{k}^{k+1}\frac{x-\lfloor x\rfloor}{x\sqrt{x^2-1}}\,dx\\\\ &=\sum_{k=1}^\infty \int_{k}^{k+1} \frac{x-k}{x\sqrt{x^2-1}}\,dx\tag1 \end{align}$$

Now, the integral under the summation sign on the right-hand side of $(1)$ can be evaluated in the closed form as

$$\begin{align} \int_{k}^{k+1} \frac{x-k}{x\sqrt{x^2-1}}\,dx&=(k+1)\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)-k\arctan\left(\frac1{\sqrt{k^2-1}}\right)\\\\ &-\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)\\\\ &+\log(\sqrt{(k+1)^2-1}+(k+1))-\log(\sqrt{k^2-1}+k) \end{align}\tag2$$

We can easily evaluate the telescoping series

$$\sum_{k=1}^\infty (k+1)\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)-k\arctan\left(\frac1{\sqrt{k^2-1}}\right)=1-\frac\pi2\tag3$$

We can also write the sum $S$ where

$$S=\sum_{k=1}^\infty \left(\log(\sqrt{(k+1)^2-1}+(k+1))-\log(\sqrt{k^2-1}+k) -\arctan\left(\frac1{\sqrt{(k+1)^2-1}}\right)\right)$$

as

$$\begin{align} S&=\lim_{K\to\infty}\left(\log(K+\sqrt{K^2-1})-\sum_{k=2}^K \arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\\\\ &=\log(2)+\lim_{K\to\infty}\left(\log(K)-\sum_{k=2}^K \arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\\\\ &=1+\log(2) -\gamma +\sum_{k=2}^\infty \left(\frac1k-\arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\tag4 \end{align}$$

Putting $(1)$-$(4)$ together, we find that

$$\int_0^{\pi/2} \bigg\{\frac{1}{\sin(x)}\bigg\}\,dx=2+\log(2)-\frac\pi2-\gamma-\sum_{k=2}^\infty \left(\frac1k-\arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\tag5$$

The series in $(5)$ converges quickly with

$$\sum_{k=2}^\infty \left(\frac1k-\arctan\left(\frac1{\sqrt{k^2-1}}\right)\right)\approx -0.0368911$$

Hence,

$$\int_0^{\pi/2} \bigg\{\frac{1}{\sin(x)}\bigg\}\,dx\approx 0.5082441$$