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 $[0,\frac{\pi}{2}]$
Canonical name	IntegratingtanXOver0fracpi2
Date of creation	2013-03-22 15:57:47
Last modified on	2013-03-22 15:57:47
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	14
Author	Wkbj79 (1863)
Entry type	Example
Classification	msc 26A42
Classification	msc 45-01
Related topic	ImproperLimits
Related topic	OneSidedLimit

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