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

Question

William Abdalla · Accepted Answer

undefined To determine if the improper integral ∫(1/(1+x^2)) dx converges or diverges from negative infinity to infinity, we use the technique of evaluating limits.

The integral can be expressed as ∫(1/(1+x^2)) dx = lim(a→-∞)∫(1/(1+x^2)) dx + lim(b→∞)∫(1/(1+x^2)) dx.

To evaluate these limits, we split the integral into two parts:

1. lim(a→-∞)∫(1/(1+x^2)) dx from -∞ to a
2. lim(b→∞)∫(1/(1+x^2)) dx from b to ∞

For the first part, lim(a→-∞)∫(1/(1+x^2)) dx from -∞ to a, we apply the Fundamental Theorem of Calculus:

∫(1/(1+x^2)) dx = arctan(x) + C

Evaluating this from -∞ to a, we get:

lim(a→-∞)[arctan(a) - arctan(-∞)]

As arctan(-∞) = -π/2, this becomes:

lim(a→-∞)[arctan(a) + π/2]

This limit does not converge, as arctan(a) grows without bound as a approaches -∞. Therefore, the integral from negative infinity to a diverges.

Similarly, for the second part, lim(b→∞)∫(1/(1+x^2)) dx from b to ∞, we also find that it diverges.

Since both parts of the integral diverge, the entire improper integral from negative infinity to infinity also diverges.

Charles Anctil · Answer

I would prove that it converges by evaluating it.

#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.