# How do you show whether the improper integral int 1/ (1+x^2) dx converges or diverges from negative infinity to infinity?

Oct 24, 2015

I would prove that it converges by evaluating it.

#### Explanation:

$\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = {\tan}^{-} 1 x + C$

If you don't know, or have forgotten the "formula", then use a trigonometric substitution:

$x = \tan \theta$ gives us $\mathrm{dx} = {\sec}^{2} \theta d \theta$ and the integral becomes

$\int \frac{1}{1 + {\tan}^{2} \theta} {\sec}^{2} \theta d \theta = \int {\sec}^{2} \frac{\theta}{\sec} ^ 2 \theta d \theta = \int d \theta = \theta + C = {\tan}^{-} 1 x + C$

Recall that ${\lim}_{x \rightarrow \infty} {\tan}^{-} 1 x = \frac{\pi}{2}$ and ${\lim}_{x \rightarrow - \infty} {\tan}^{-} 1 x = - \frac{\pi}{2}$

We need to split the integral. $0$ is usually easy to work with, so let's use it.

${\int}_{-} {\infty}^{\infty} \frac{1}{1 + {x}^{2}} \mathrm{dx} = {\lim}_{a \rightarrow - \infty} {\int}_{a}^{0} \frac{1}{1 + {x}^{2}} \mathrm{dx} + {\lim}_{b \rightarrow \infty} {\int}_{0}^{\infty} \frac{1}{1 + {x}^{2}} \mathrm{dx}$

$= {\lim}_{a \rightarrow - \infty} {\left[{\tan}^{-} 1 x\right]}_{a}^{0} +$ ${\lim}_{b \rightarrow \infty} {\left[{\tan}^{-} 1 x\right]}_{0}^{b}$

$= {\lim}_{a \rightarrow - \infty} \left[{\tan}^{-} 1 \left(0\right) - {\tan}^{-} 1 \left(b\right)\right] + {\lim}_{b \rightarrow \infty} \left[{\tan}^{-} 1 b - {\tan}^{-} 1 0\right]$

$= {\lim}_{a \rightarrow - \infty} \left[0 - {\tan}^{-} 1 \left(b\right)\right] + {\lim}_{b \rightarrow \infty} \left[{\tan}^{-} 1 b - 0\right]$

$= \left[- \left(- \frac{\pi}{2}\right)\right] + \left[\frac{\pi}{2}\right] = \pi$

The integral converges.