\begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx & =\int_{0}^{\tan^{\bullet}\infty}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx\\ & =\sum_{k=0}^{\infty}\int_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\frac{\left\lfloor \tan x\right\rfloor }{\tan x}dx\\ & =\sum_{k=0}^{\infty}\int_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\frac{k}{\tan x}dx\\ & =\sum_{k=1}^{\infty}k\int_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\frac{1}{\tan x}dx\\ & =\sum_{k=1}^{\infty}k\left[\log\left(\sin x\right)\right]_{\tan^{\bullet}k}^{\tan^{\bullet}\left(k+1\right)}\\ & =\sum_{k=1}^{\infty}k\left\{ \log\left(\sin\tan^{\bullet}\left(k+1\right)\right)-\log\left(\sin\tan^{\bullet}\left(k\right)\right)\right\} \\ & =\sum_{k=1}^{\infty}k\left\{ \log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-\log\frac{k}{\sqrt{1+k^{2}}}\right\} \cmt{\because\sin\tan^{\bullet}z=\frac{z}{\sqrt{1+z^{2}}}}\\ & =\sum_{k=1}^{\infty}\left\{ \left(k+1\right)\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-k\log\frac{k}{\sqrt{1+k^{2}}}\right\} \\ & =\sum_{k=1}^{\infty}\left\{ \left(k+1\right)\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}-k\log\frac{k}{\sqrt{1+k^{2}}}\right\} -\sum_{k=1}^{\infty}\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}\\ & =-\log\frac{1}{\sqrt{2}}-\sum_{k=1}^{\infty}\log\frac{k+1}{\sqrt{1+\left(k+1\right)^{2}}}\\ & =\frac{1}{2}\log2-\frac{1}{2}\sum_{k=1}^{\infty}\log\frac{\left(k+1\right)^{2}}{1+\left(k+1\right)^{2}}\\ & =\frac{1}{2}\log2+\frac{1}{2}\sum_{k=1}^{\infty}\log\frac{\left(k+1\right)^{2}+1}{\left(k+1\right)^{2}}\\ & =\frac{1}{2}\log2+\frac{1}{2}\sum_{k=1}^{\infty}\log\left(1+\frac{1}{\left(k+1\right)^{2}}\right)\\ & =\frac{1}{2}\log2+\frac{1}{2}\log\prod_{k=1}^{\infty}\left(1+\frac{1}{\left(k+1\right)^{2}}\right)\\ & =\frac{1}{2}\log2+\frac{1}{2}\log\prod_{k=2}^{\infty}\left(1+\frac{1}{k^{2}}\right)\\ \\ & =\frac{1}{2}\log2+\frac{1}{2}\log\left(\frac{1}{2}\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)\right)\\ & =\frac{1}{2}\log2-\frac{1}{2}\log2+\frac{1}{2}\log\left(\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)\right)\\ & =\frac{1}{2}\log\left(\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)\right)\\ & =\frac{1}{2}\log\frac{\sinh\pi}{\pi}\cmt{\because\pi z\prod_{k=1}^{\infty}\left(1+\frac{z}{k^{2}}\right)=\sinh\left(\pi z\right)} \end{align*}