# Testing for converging integrals

• Oct 28th 2010, 07:50 PM
Glitch
Testing for converging integrals
The question:

Use the inequality form of the comparison test to determine whether or not the following improper integral converges

$\displaystyle \int_{1}^{\infty} \frac{1}{\sqrt{1 + x^4}} dx$

My attempt:
When x is large, dominant term is $\displaystyle x^4$

$\displaystyle \frac{1}{\sqrt{1 + x^4}} < \frac{1}{\sqrt{x^4}}$

$\displaystyle \int_{1}^{\infty} \frac{1}{x^2} dx$

$\displaystyle _{1}^{\infty}[\frac{-1}{x}]$

As x -> infinity, function goes to 0:
0 - (-1)

So the integral converges (approaches 1). Is this correct? Thanks.
• Oct 29th 2010, 12:07 AM
Glitch
Fixed my previous mistake. Still wondering if this is correct.
• Oct 29th 2010, 07:50 AM
Krizalid
no need to compute the last integral since you know integrals of the form $\displaystyle \displaystyle\int_1^\infty\frac{dx}{x^p},\,p>1$ converge.
• Oct 29th 2010, 08:06 AM
Also sprach Zarathustra
$\displaystyle \displaystyle\int_{1}^{\infty} \frac{1}{\sqrt{1 + x^4}} dx=\frac{2 \Gamma(\frac{5}{4})^2}{\sqrt\pi}$