Motivated by Calculate $\int_0^\infty \frac{\{\tan x\}}{\tan x}dx$
I want to evaluate the following integral,
$$I=\int_{0}^{\pi/2}\frac{ \{ \tan x \} }{\tan x}dx$$
Here is the value from Wolfram,
$$I=0.919900494885649087058609882333342085980315192991199436742670023056103944536683559170770204$$
First step would be to write,
$$\{ \tan x \} =\tan x-\lfloor\tan x\rfloor$$
$$I=\frac{\pi}{2}-\int_{0}^{\frac{\pi}{2}}\frac{\lfloor\tan x\rfloor}{\tan x}dx$$
$$x\to \pi/2-x$$
$$I=\frac{\pi}{2}-\int_{0}^{\frac{\pi}{2}}\frac{\lfloor\cot x\rfloor}{\cot x}dx$$
$$x\to \cot x$$
$$I=\frac{\pi}{2}-\int_{0}^{\infty}\frac{\operatorname{floor}\left(x\right)}{x\csc^{2}\left(\operatorname{arccot}x\right)}dx$$
$$I=\frac{\pi}{2}-\int_{0}^{\infty}\frac{\operatorname{floor}\left(x\right)}{x\left(1+x^{2}\right)}dx$$
$$I=\frac{\pi}{2}-\sum_{j=1}^{\infty}j\left(\int_{j}^{1+j}\frac{1}{x\left(1+x^{2}\right)}dx\right)$$
Also,
$$\int_{j}^{1+j}\frac{1}{x\left(1+x^{2}\right)}dx=\ln\left(\frac{1+j}{j}\right)-\frac{1}{2}\ln\left(\frac{\left(j+1\right)^{2}+1}{j^{2}+1}\right)$$
or,
$$=\ln\left(\frac{1+j}{j}\left(\frac{j^{2}+1}{\left(j+1\right)^{2}+1}\right)^{\frac{1}{2}}\right)$$
Therefore,
$$I=\frac{\pi}{2}-\sum_{j=1}^{\infty}j\ln\left(\frac{1+j}{j}\left(\frac{j^{2}+1}{\left(j+1\right)^{2}+1}\right)^{\frac{1}{2}}\right)$$
or,
$$I=\frac{\pi}{2}-\ln\left(\prod_{j=1}^{\infty}\left(\frac{1+j}{j}\right)^{j}\left(\frac{j^{2}+1}{\left(j+1\right)^{2}+1}\right)^{\frac{j}{2}}\right)$$
Note that,
$$\prod_{j=1}^{n}\left(\frac{1+j}{j}\right)^{j}=\frac{\left(n+1\right)^{n}}{n!}$$
$$\prod_{j=1}^{n}\left(\frac{j^{2}+1}{\left(j+1\right)^{2}+1}\right)^{\frac{j}{2}}=\frac{1}{\left(\left(n+1\right)^{2}+1\right)^{\frac{n}{2}}}\prod_{j=1}^{n}\sqrt{1+j^{2}}$$
$$\prod_{j=1}^{\infty}\left(\frac{1+j}{j}\right)^{j}\left(\frac{j^{2}+1}{\left(j+1\right)^{2}+1}\right)^{\frac{j}{2}}=\lim_{n\to \infty}\left(\frac{\left(n+1\right)^{n}}{n!}\cdot\frac{1}{\left(\left(n+1\right)^{2}+1\right)^{\frac{n}{2}}}\prod_{j=1}^{n}\sqrt{1+j^{2}}\right)$$
$$=\lim_{n\to \infty}\left(\frac{\left(n+1\right)^{n}}{\left(\left(n+1\right)^{2}+1\right)^{\frac{n}{2}}}\right)\cdot\lim_{n\to \infty}\left(\frac{\prod_{j=1}^{n}\sqrt{1+j^{2}}}{n!}\right)$$
Although I am not aware of the proof but it seems that,
$$\lim_{n\to \infty}\left(\frac{\prod_{j=1}^{n}\sqrt{1+j^{2}}}{n!}\right)=\sqrt{\frac{\sinh\pi}{\pi}}$$
whereas,
$$\lim_{n\to \infty}\left(\frac{\left(n+1\right)^{n}}{\left(\left(n+1\right)^{2}+1\right)^{\frac{n}{2}}}\right)=1$$
This would imply,
$$\boxed{I=\frac{\pi}{2}-\frac{1}{2}\ln\left(\frac{\sinh\pi}{\pi}\right)}$$
But this does not work numerically.
Although I have not done this rigorously, I can't seem to find where the mistake is.
Best Answer
Your approach is fine and my numerical simulations confirm it up to 3 decimal digits. It is also correct that $$ \lim_{n\to \infty}\left(\frac{\prod_{j=1}^{n}\sqrt{1+j^{2}}}{n!}\right)=\prod_{j=1}^\infty \sqrt{1+\frac{1}{j^2}}=\sqrt{\frac{\sinh\pi}{\pi}}, $$ or similarly $$ \prod_{j=1}^\infty \left(1+\frac{1}{j^2}\right)=\frac{\sinh\pi}{\pi}. $$ In the following, we will prove the latter result to conclude your solution.
Denote $A=\prod_{j=1}^\infty \left(1+\frac{1}{j^2}\right)$. Therefore, $$\ln A{=\sum_{j=1}^\infty \ln\left(1+\frac{1}{j^2}\right) \\= \sum_{j=1}^\infty \sum_{n=1}^\infty\frac{(-1)^{n+1}}{nj^{2n}} \\= \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\zeta(2n). } $$ Now, by replacing $\zeta(2n)=\frac{(-1)^{n+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$, where $B_n$ is the Bernoulli number, we obtain $$ \ln A {= \sum_{n=1}^\infty\frac{(2\pi)^{2n}B_{2n}}{2n(2n)!} \\= \sum_{n=1}^\infty\frac{(2\pi)^{2n}B_{2n}\left[1+(-1)^{2n}\right]}{4n(2n)!} \\= \sum_{n=1}^\infty\frac{(2\pi)^{n}B_{n}\left[1+(-1)^{n}\right]}{2n(n)!} \\= \sum_{n=1}^\infty\frac{(2\pi)^{n}B_{n}}{2n(n)!} + \sum_{n=1}^\infty\frac{(-2\pi)^{n}B_{n}}{2n(n)!}. } $$ The characteristics function of Bernoulli numbers is given by $$ \frac{x}{e^x-1}=\sum_{n=0}^\infty \frac{B_nx^n}{n!}, $$ from which, the following deductions are made: $$ \frac{x}{e^x-1}=1+\sum_{n=1}^\infty \frac{B_nx^n}{n!} {\implies \frac{1}{e^x-1}=\frac{1}{x}+\sum_{n=1}^\infty \frac{B_nx^{n-1}}{n!} \\\implies \frac{1}{e^x-1}-\frac{1}{x}=\sum_{n=1}^\infty \frac{B_nx^{n-1}}{n!} \\\overset{\int}{\implies} \ln\left|\frac{e^{-x}-1}{x}\right|=\sum_{n=1}^\infty \frac{B_nx^n}{n\cdot n!}. } $$ Therefore $$ { \sum_{n=1}^\infty \frac{B_n(2\pi)^n}{n\cdot n!}=\ln\left|\frac{e^{-2\pi}-1}{2\pi}\right| \\ \sum_{n=1}^\infty \frac{B_n(-2\pi)^n}{n\cdot n!}=\ln\left|\frac{e^{2\pi}-1}{2\pi}\right|, } $$ and we finally arrive at $$ \ln A{= \sum_{n=1}^\infty\frac{(2\pi)^{n}B_{n}}{2n(n)!} + \sum_{n=1}^\infty\frac{(-2\pi)^{n}B_{n}}{2n(n)!} \\=\frac{1}{2}\ln\left|\frac{e^{-2\pi}-1}{2\pi}\right|+\frac{1}{2}\ln\left|\frac{e^{2\pi}-1}{2\pi}\right| \\=\frac{1}{2}\ln\left|\frac{2-e^{-2\pi}-e^{2\pi}}{4\pi^2}\right| \\=\frac{1}{2}\ln\left|\frac{2-2\cosh 2\pi}{4\pi^2}\right| \\=\frac{1}{2}\ln\left|\frac{4\sinh^2\pi}{4\pi^2}\right| \\=\ln \frac{\sinh\pi}{\pi}. } $$ Therefore, $A=\prod_{j=1}^\infty \left(1+\frac{1}{j^2}\right)=\frac{\sinh\pi}{\pi}$ $\blacksquare$