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

1 Answer
Oct 24, 2015

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.