Range of $\ln\tan^{-1}x$

Question

I can't believe I am asking such a silly question. So I have the function $$\ln\tan^{-1}x$$ I am asked to find the range of this function. I know that the range of $\ln x$ is all real numbers and that the range of $\tan^{-1}(x)$ is $(-\frac\pi2$, $\frac\pi2)$. Wouldn't the range of $\ln\tan^{-1}x$ also be $(-\frac\pi2$, $\frac\pi2)$, or am I doing something really stupid?

Answer 1

The range of $\ln\tan^{-1}x$ is the range of $\ln y$ where $y\in(-\frac\pi2,\frac\pi2)$.

But wait a minute: the logarithm accepts only positive real numbers. Hence we can restrict $y$ to $(0,\frac\pi2)$, since the natural logarithm will be undefined for arguments outside this range. Then the range of $\ln\tan^{-1}x$ is $(-\infty,\ln\frac\pi2)$: $\ln 0$ is "$-\infty$", $\ln\frac\pi2$ needs no further simplification and the natural logarithm is a strictly increasing function.

Answer 2

Note that in order for $\ln \tan^{-1}(x)$ to be defined, one must have $\tan^{-1}(x) > 0$, i.e. $x > 0$.

First we find the range of $\tan^{-1}(x)$ for $x \in {]0, + \infty[}$. Since $\tan^{-1}$ is continuous and monotonic, the range is the interval $]0, \frac{\pi}{2} [$, because $\lim_{x \to 0^+} \tan^{-1}(x) = 0$ and $\lim_{x \to +\infty} \tan^{-1}(x) = \frac{\pi}{2}$.

Then, we find the range of $\ln x$ for $x \in {]0, \frac{\pi}{2}[}$. Since $\ln$ is continuous and monotonic, the range is the interval $]-\infty, \ln \frac{\pi}{2}[$, because $\lim_{x \to 0^+} \ln x = -\infty$ and $\lim_{x \to \frac{\pi}{2}^-} \ln x = \ln \frac{\pi}{2}$.

Thus, the range of $\ln \tan^{-1}(x)$ is $]-\infty, \ln \frac{\pi}{2}[$.