# Comparison Theorem for integrals

1. Mar 2, 2004

### Corky

I have to use the comparison theorem to find if the function converges or diverges. Any ideas as to what I can use to compare the function to??

integrate: sqrt(1 + sqrt(x))/sqrt(x)dx
Using the comparison theorem!

2. Mar 2, 2004

### Tom Mattson

Staff Emeritus
You need a function g(x) for which your function (call it f(x)) satisfies the following:

limx-->&infin;(f/g)=k

where k is a positive real number.

Look at the leading terms in the numerator and denominator of f(x). Rewriting your function f(x) in radical form, it looks like.

f(x)=(x1/2+1)/x1/2

For large x, we can ignore the "+1" in the numerator to get:

f(x)~x1/4/x1/2=x-1/4

g(x) looks to me like a good choice for your test function. Why? Because it is easy to establish the convergence of that integral (I assume you are integrating from 1 to infinity or something like that). Try the limit comparison test on that and see what happens.

Also, your original function is easy to integrate, so you can check your answer. Just let u=x1/2 and so du=(1/2)x-1/2.

edit: fixed HTML code

3. Mar 2, 2004

### Corky

(1 + x^1/2)/x <= ((1 + x^1/2)/x)^1/2 for all x >= 1.
Thus since the integral of(1 + x^1/2)/x diverges as does the original function.
Thanks