Question

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

Answer 1

I would prove that it converges by evaluating it.

Explanation:

`int 1/(1+x^2) dx = tan^-1x +C`

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

`x = tan theta` gives us `dx = sec^2 theta d theta` and the integral becomes

`int 1/(1+tan^2theta) sec^2theta d theta = int sec^2theta/sec^2theta d theta = int d theta = theta +C= tan^-1 x +C`

Recall that `lim_(xrarroo)tan^-1x = pi/2` and `lim_(xrarr-oo)tan^-1x = -pi/2`

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

`int_-oo^oo 1/(1+x^2) dx = lim_(ararr-oo) int_a^0 1/(1+x^2) dx + lim_(brarroo) int_0^oo 1/(1+x^2) dx`

`= lim_(ararr-oo) [tan^-1 x]_a^0 +` `lim_(brarroo) [tan^-1 x]_0^b`

`= lim_(ararr-oo)[tan^-1(0) - tan^-1(b)] + lim_(brarroo)[tan^-1b-tan^-1 0]`

`= lim_(ararr-oo)[0 - tan^-1(b)] + lim_(brarroo)[tan^-1b-0]`

`= [-(-pi/2)]+[pi/2] = pi`

The integral converges.