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Here is a solution I read when trying to solve a problem, and I can't figure out how it jumped in this step here:

$$ \frac{1}{x}\left(\frac{\pi}{2} - \arctan\frac{1}{x}\right) = \frac{1}{x} \arctan x $$

This was related to a limit and integral problem where $x \to 0^+$. Please let me know if more information are needed and I will edit!

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user321070
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    Use http://math.stackexchange.com/questions/304399/are-mathrmarccotx-and-arctan1-x-the-same-function – lab bhattacharjee May 29 '16 at 06:22
  • With the usual convention that $\tan^{-1}x\in(-\frac{\pi}{2},\frac{\pi}{2})$ we have $\tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{\pi}{2}$ for $x>0$ and $-\frac{\pi}{2}$ for $x<0$. – almagest May 29 '16 at 06:32

1 Answers1

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I assume you mean this holds for all $x>0$.

  • The factor $\frac{1}{x}$ on both sides is not really helping, since for any $x\neq 0$ your statement is equivalent to $$ \frac{\pi}{2} - \arctan \frac{1}{x} = \arctan x, \qquad x > 0 \tag{1} $$
  • Now, rearrange the terms: (1) becomes equivalent to $$ \frac{\pi}{2} = \arctan \frac{1}{x} + \arctan x, \qquad x > 0 \tag{2} $$

which is a known identity.

One way (maybe not the most elegant) to prove this last identity is to observe that the function $f\colon (0,\infty)\to \mathbb{R}$ defined by $f(x) = \arctan \frac{1}{x} + \arctan x$ is differentiable, and (using the derivative $\arctan' x = \frac{1}{1+x^2}$) that $f'(x) = 0$ for all $x>0$. So $f$ is constant, and since $$\lim_{x\to\infty} f(x) = \arctan 0 + \lim_{x\to\infty}\arctan x = \lim_{x\to\infty}\arctan x = \frac{\pi}{2}$$ you get the result.

Clement C.
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  • Thanks for replying. The step I posted was near the final solution, and the $ \frac{1}{x} $ was needed. The final answer given was this: $ \lim_{x \to 0^+} \frac{\arctan x}{x} = 1 $. So I was trying to understand that step I posted in the question. – user321070 May 29 '16 at 06:34
  • @user321070 I'm not saying the $1/x$ should not have been there in your proof (which I do not know); simply that to establish the equality you need, getting rid of it can be done safely, and helps. – Clement C. May 29 '16 at 06:35
  • Ah ok, I understand now. It was strange and out of blue since this identity you mentioned has never occurred in my study until now. Thanks for helping out! – user321070 May 29 '16 at 06:42
  • You're welcome! – Clement C. May 29 '16 at 06:42