# integrating $\mathop{tan}\nolimits x$ over $[0,\frac{\pi}{2}]$

Note that what is meant by $\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\tan x\,dx$ is actually $\displaystyle\lim_{t\to\frac{\pi}{2}^{-}}\int\limits_{0}^{t}\tan x\,dx$, since $\tan x$ is defined on $[0,\frac{\pi}{2})$ but not at $\frac{\pi}{2}$.

$\begin{array}[]{ll}\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\tan x\,dx&% \displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\int\limits_{0}^{t}\tan x\,dx\\ &\\ &\displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\ln|\sec x|\bigg{|}_{0}^{t}\\ &\\ &\displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\ln|\sec t|-\ln|\sec 0|\\ &\\ &\displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\ln|\sec t|\\ &\\ &\displaystyle=\infty.\end{array}$

Title integrating $\mathop{tan}\nolimits x$ over $[0,\frac{\pi}{2}]$ IntegratingtanXOver0fracpi2 2013-03-22 15:57:47 2013-03-22 15:57:47 Wkbj79 (1863) Wkbj79 (1863) 14 Wkbj79 (1863) Example msc 26A42 msc 45-01 ImproperLimits OneSidedLimit