Evaluate the following integral: $\int_0^{\frac{\pi}{2}} \lbrace\tan x\rbrace\mathrm{d}x$

Question

$$I=\int_0^{\frac{\pi}{2}} \lbrace\tan x\rbrace\mathrm{d}x$$ So I have this interesting integral and I tried to evaluate it:

Let $u=\tan x$ we'll have the following: \begin{align} \int_0^\infty \frac{\lbrace u\rbrace}{1+u^2 }\mathrm{d}u&=\int_0^\infty \frac{u-\lfloor u\rfloor}{1+u^2}\mathrm{d}u\\ &=\lim_{a\to +\infty} \Bigg(\int_0^a \frac{u}{u^2+1}\mathrm{d}u-\int_0^a\frac{\lfloor u\rfloor}{u^2+1}\mathrm{d}u\Bigg) \end{align} And the first part is obvious: $$\int_0^a \frac{u}{u^2+1}\mathrm{d}u=\frac{\ln (a^2+1)}2$$ Therefore: $$I=\lim_{a\to +\infty}\Bigg(\frac{\ln (a^2+1)}2-\underbrace{\int_0^a\frac{\lfloor u\rfloor}{u^2+1}\mathrm{d}u}_{F}\Bigg)$$ And $F$ is the part where I got stuck because $\lfloor x\rfloor$ can't be simplified or approximated. Any thoughts or hints?

Answer 1

Note that we can write

$$\begin{align} \int_{0}^{N}\frac{x-\lfloor x\rfloor }{x^2+1}\,dx&=\frac12\log(N^2+1)-\sum_{n=1}^N\int_{n-1}^n \frac{\lfloor x\rfloor }{x^2+1}\,dx\\\\ &=\frac12\log(N^2+1)-\sum_{n=1}^N (n-1)(\arctan(n)-\arctan(n-1))\\\\ &=\frac12\log(N^2+1)+\sum_{k=1}^N \arctan(n)-N\arctan(N)\\\\ &=\frac12\log(N^2+1)+N(\arctan(1/N))-\sum_{n=1}^N \arctan(1/n)\\\\ &=\log(N)+1-\sum_{n=1}^N \arctan(1/n)+O\left(\frac{1}{N^2}\right)\\\\ &=\sum_{n=1}^N \left(\frac1n-\arctan(1/n)\right) -\gamma +1+O\left(\frac1N\right) \end{align}$$

Letting $N\to\infty$, we find that

$$\int_0^{\pi/2}\tan(\{x\})\,dx=1-\gamma +\sum_{n=1}^\infty \left(\frac1n-\arctan(1/n)\right)$$

Answer 2

This got messy towards the end.

$\begin{array}\\ I &=\int_0^{\frac{\pi}{2}} \{\tan x\}dx\\ &=\int_0^{\frac{\pi}{4}} \{\tan x\}dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \{\tan x\}dx\\ &=\int_0^{\frac{\pi}{4}} \tan xdx+\int_0^{\frac{\pi}{4}} \{\tan (\pi/2-x)\}dx\\ &=\dfrac{\ln(2)}{2}+\int_0^{\frac{\pi}{4}} \{\dfrac1{\tan (x)}\}dx\\ &=\dfrac{\ln(2)}{2}+\sum_{k=1}^{\infty}\int_{\tan^{(-1)}(1/(k+1))}^{\tan^{(-1)}(1/k)} \{\dfrac1{\tan (x)}\}dx \qquad (*)\\ &=\dfrac{\ln(2)}{2}+\sum_{k=1}^{\infty}\int_{v_{k+1}}^{v_k} \{\dfrac1{\tan (x)}\}dx \qquad v_k = \tan^{(-1)}(1/k)\\ &=\dfrac{\ln(2)}{2}+\sum_{k=1}^{\infty}\int_{v_{k+1}}^{v_k} (\dfrac1{\tan (x)}-k)dx\\ &=\dfrac{\ln(2)}{2}+\sum_{k=1}^{\infty}\left(\int_{v_{k+1}}^{v_k} \dfrac1{\tan (x)}dx-\int_{v_{k+1}}^{v_k} kdx\right)\\ &=\dfrac{\ln(2)}{2}+\sum_{k=1}^{\infty}\left(\ln(\sin(t))|_{v_{k+1}}^{v_k}-k(v_k-v_{k+1})\right) \qquad (**)\\ &=\dfrac{\ln(2)}{2}+\sum_{k=1}^{\infty}\left(\ln(\dfrac{1}{\sqrt{k^2+1}})-\ln(\dfrac{1}{\sqrt{(k+1)^2+1}})-k(v_k-v_{k+1})\right)\\ &=\dfrac{\ln(2)}{2}+\sum_{k=1}^{\infty}\left(\ln(\dfrac{\sqrt{(k+1)^2+1}}{\sqrt{k^2+1}})-k(v_k-v_{k+1})\right)\\ &=\dfrac{\ln(2)}{2} +\lim_{m \to \infty}\sum_{k=1}^{m}\left(\ln(\dfrac{\sqrt{(k+1)^2+1}}{\sqrt{k^2+1}})-k(v_k-v_{k+1})\right)\\ &=\dfrac{\ln(2)}{2} +\lim_{m \to \infty}\left((\ln(\dfrac{\sqrt{(m+1)^2+1}}{\sqrt{2}})-\sum_{k=1}^{m}k(v_k-v_{k+1})\right)\\ &=\dfrac{\ln(2)}{2} +\lim_{m \to \infty}\left((\ln(\dfrac{\sqrt{(m+1)^2+1}}{\sqrt{2}})-\sum_{k=1}^{m}kv_k+\sum_{k=1}^{m}kv_{k+1})\right)\\ &=\dfrac{\ln(2)}{2} +\lim_{m \to \infty}\left((\ln(\dfrac{\sqrt{(m+1)^2+1}}{\sqrt{2}})-\sum_{k=1}^{m}kv_k+\sum_{k=2}^{m+1}(k-1)v_{k})\right)\\ &=\dfrac{\ln(2)}{2} +\lim_{m \to \infty}\left((\ln(\dfrac{\sqrt{(m+1)^2+1}}{\sqrt{2}})-\sum_{k=1}^{m}kv_k+\sum_{k=2}^{m+1}kv_{k}-\sum_{k=2}^{m+1}v_{k})\right)\\ &=\dfrac{\ln(2)}{2} +\lim_{m \to \infty}\left(\frac12(\ln(m^2+2m+2)-\ln(2))-v_1+(m+1)v_{m+1}-\sum_{k=2}^{m+1}v_{k})\right)\\ &=\lim_{m \to \infty}\left(\frac12\ln(m^2+2m+2)+(m+1)v_{m+1}-\sum_{k=1}^{m+1}v_{k})\right)\\ &=1+\lim_{m \to \infty}\left(\frac12(\ln(m^2)+\ln(1+2/m+2/m^2)-\sum_{k=1}^{m+1}v_{k})\right)\\ &=1+\lim_{m \to \infty}\left(\ln(m)-\sum_{k=1}^{m+1}v_{k})\right)\\ &=1+\lim_{m \to \infty}\left(\ln(m)-\sum_{k=1}^{m+1}\tan^{(-1)}(1/k)\right)\\ \end{array} $

(*) Want $k \le \dfrac1{\tan(x)} \le k+1 $ so $1/k \le \tan(x) \le 1/(k+1) $ so $\tan^{(-1)}(1/(k+1)) \le x \le \tan^{(-1)}(1/k) $

(**) Since $\sin(\arctan(x)) =\dfrac{x}{\sqrt{x^2+1}} $, $\sin(v_k) =\sin(\tan^{(-1)}(1/k)) =\dfrac{1/k}{\sqrt{(1/k)^2+1}} =\dfrac{1}{\sqrt{k^2+1}} $

$v_k-v_{k+1} =\tan^{(-1)}(1/k)-\tan^{(-1)}(1/(k+1)) =\tan^{(-1)}(\dfrac{\frac1{k}-\frac1{k+1}}{1+\frac1{k(k+1)}}) =\tan^{(-1)}(\dfrac{1}{k(k+1)+1}) $