How does $ \frac{1}{x}\left(\frac{\pi}{2} - \arctan\frac{1}{x}\right)$ simplify to $\frac{1}{x} \arctan x $?

Question

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!

Answer 1

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.

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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.