# How do you find the integral of int (1/(1+x^2) ) dx from negative infinity to 0?

Sep 20, 2015

$\frac{\pi}{2}$

#### Explanation:

let $x = \tan \theta$
$\mathrm{dx} = {\sec}^{2} \theta d \left(\theta\right)$
$1 + {\tan}^{2} \theta = {\sec}^{2} \theta$
$\mathmr{if} x = - \infty \theta = - \frac{\pi}{2}$
$\mathmr{if} x = 9 \theta = 0$
${\int}_{-} {\infty}^{01} / \left(1 + {x}^{2}\right) \mathrm{dx} = {\int}_{-} {\left(\frac{\pi}{2}\right)}^{0} {\sec}^{2} \frac{\theta}{\sec} ^ 2 \theta d \left(\theta\right)$
=${\int}_{-} {\left(\frac{\pi}{2}\right)}^{0} d \left(\theta\right)$=${\left[\theta\right]}_{- \frac{\pi}{2}}^{0} = 0 - \left(- \frac{\pi}{2}\right) = \frac{\pi}{2}$

Sep 21, 2015

This is related to $\arctan x$...

$\frac{d}{\mathrm{dx}} \left[\arctan x\right] = \frac{1}{1 + {x}^{2}}$

$\therefore \text{ } {\int}_{a}^{b} \frac{1}{1 + {x}^{2}} \mathrm{dx} = \arctan x + C$

${\int}_{- \infty}^{0} \frac{1}{1 + {x}^{2}} = \arctan \left(0\right) - \arctan \left(- \infty\right)$

Due to $\tan x$ having a domain of $\left(\frac{\pi}{2} , \frac{\pi}{2}\right) \pm \pi k$ where $k \in \mathbb{Z}$, the inverse, $\arctan x$, is a graph with a range of $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, and it looks kind of like a sideways ${x}^{3}$, crossing through $\left(0 , 0\right)$.

graph{arctanx [-20, 20, -3.12, 3.12]}

Therefore, as $x \to - \infty$, you get $- \frac{\pi}{2}$, and so:

$= 0 - \left(- \frac{\pi}{2}\right) = \textcolor{b l u e}{\frac{\pi}{2}}$