Show limit of $\arctan\left(\frac{x}{\pi}\right)$ as $x\rightarrow\infty$

Question

I have been given the question $$\int^{\infty}_{0}\frac{dx}{x^2+\pi^2}=\frac{1}{2}$$

I have found the integral and obtained $$\int^{\infty}_{0}\frac{dx}{x^2+\pi^2} = \lim_{N\rightarrow\infty}\int^{N}_{0}\frac{dx}{x^2+\pi^2}=\frac{1}{\pi}\lim_{N\rightarrow\infty}\left[\arctan\left(\frac{N}{\pi}\right)\right] $$ and came to sub in my limits of integration and I am unable to show that this limit does equal $\frac{\pi}{2}$ which would allow me to show the question is true. I need a full method of how to do it rather than saying "since the integral is = to $\frac{1}{2} $ then it must be $\frac{\pi}{2}$.

Answer 1

Recall that by definition for the inverse of $\tan x$ we have

$$\arctan x: \mathbb{R}\to \left(-\frac \pi 2,\frac \pi 2\right)$$

with $\arctan x$ strictly increasing.

Answer 2

Hint : Let $\frac{x}{\pi}=\tan (t)$ where $t\in (-\pi/2,\pi/2)$. When does $\tan(t)$ go to $+\infty$ ?

Answer 3

Recall the geometric meaning of $\tan^{-1}x$, as in the figure below. i.e. the angle, $\theta$, subtended by the line segment, $PQ=\tan\theta=x$, that is tangent to the unit circle at $P$. As user points out, this angle is assumed, by definition, to be the principal value, in $(-\frac{\pi}{2},\frac{\pi}{2})$.

When $Q$ is raised infinitely high, where does the intersection between $OQ$ and $PQ$ move to? And what would that imply about the lines $OQ$ and $PQ$? From this, you can infer what happens to $\theta=\tan^{-1}x$ as $x\to\infty$. This can then show you the range of $\tan^{-1}x$.