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$$\int_{-1}^{1}\tan\left(\frac1x\right) dx$$ How do I proceed? Please help.

Robert Z
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Najmul
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    How do you even know that the function is integrable? I do not know which integral(Riemann/Lebesgue) you are dealing with, but this function is unbounded at various points in the given domain. If it happens to somehow exist, then it can only be $0$, because the function $\tan \frac 1x$ is an odd function wherever defined. – Sarvesh Ravichandran Iyer Mar 03 '18 at 08:25
  • why is this $$\partial x$$? and not $dx$ – Dr. Sonnhard Graubner Mar 03 '18 at 08:37
  • This integral has a Cauchy principal value, which is $0. \qquad$ – Michael Hardy Mar 08 '18 at 02:31
  • Please remember that you can choose an aswer among the given is the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 09 '18 at 22:55

2 Answers2

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Hint. The function $\tan(1/x)$ is not integrable in a neighbourhood of $\frac{2}{(2k+1)\pi}\in [-1,1]$ for all $k\in\mathbb{Z}$ because as $x\to \frac{2}{(2k+1)\pi}$ $$\tan(1/x)\sim \frac{C_k}{x-\frac{2}{(2k+1)\pi}}.$$

Robert Z
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Note that for $x=\frac{2}{\pi}+h$ with $h\to 0$

$$\tan\frac1x=\tan\frac1{\frac{2}{\pi}+h}=\tan\left[\left(\frac{2}{\pi}+h\right)^{-1}\right]=\\\tan\left[\frac{\pi}2\left(1+\frac{\pi h}{2}\right)^{-1}\right]=\tan\left[\frac{\pi}2\left(1-\frac{\pi h}{2}+o(h)\right)\right]=\tan{\left(\frac{\pi}2-\frac{\pi^2 h}{4}+o(h)\right)}=\frac1{\tan{\left(\frac{\pi^2 h}{4}+o(h)\right)}}\sim\frac1{\frac{\pi^2 h}{4}}$$

and thus the integral does not converge.

Michael Hardy
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user
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