How do you show whether the improper integral $\int \frac{1}{1 + x^{2}} d x$ $int 1/ (1+x^2) dx$ converges or diverges from negative infinity to infinity?

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How do you show whether the improper integral $\int \frac{1}{1 + x^{2}} d x$ $int 1/ (1+x^2) dx$ converges or diverges from negative infinity to infinity?

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Answer 1 · 2015-10-24T17:28:04

I would prove that it converges by evaluating it.

Explanation:

$\int \frac{1}{1 + x^{2}} d x = {tan}^{- 1} x + C$ $int 1/(1+x^2) dx = tan^-1x +C$

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

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

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

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.

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How do you show whether the improper integral $\int \frac{1}{1 + x^{2}} d x$ $int 1/ (1+x^2) dx$ converges or diverges from negative infinity to infinity?

1 Answer

Explanation:

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How do you show whether the improper integral $\int \frac{1}{1 + x^{2}} d x$ $int 1/ (1+x^2) dx$ converges or diverges from negative infinity to infinity?