Question

When using comparison theorem to determine whether an integral is convergent or divergent how do you know which function to use. For example, one specific questions asks x/sqrt(1+x^6) between 1 and infiniti (lets call this g(x)). Now in order to use comparison theory we have to find a function, f(x) which is greater or equal to g(x). Can I let f(x) =x which will always be greater than g(x)? Or is there some sort of trick to figure out which function to use?

Answer 1

There is no special trick. Just like the comparison tests for infinity series. If you have a series:
\[\sum_{k=1}^{\infty}\frac{1}{n}\]
You can compare it to:
\[\sum_{k=1}^{\infty}\frac{2}{n}\] because its just a tiny bit bigger. (It may not be practical)
So x is always bigger. You know that integral converges. That means a smaller integral must also converge.

Answer 2

should read "infinite series." in the first line.

Answer 3

\[\lim_{t \rightarrow \inf}\int\limits_{1}^{t}xdx = (x^2)/2\ =\lim_{t \rightarrow \inf} (t^2)/2) -1/2\] This would be divergent? which would mean that g(x) would also be divergent correct?

Answer 4

If a larger series diverges, that says nothing about a smaller series :P You would have to find a smaller integral that diverges to show that a larger integral diverges.

Answer 5

Try this and go down to comparison tests. http://science.kennesaw.edu/~plaval/math2202/intimp.pdf

Answer 6

I deleted the first one because it wasn't the right link.

Answer 7

Well, I am confused the theorem in my book says if \[f(x) \ge g(x)\] and f(x) either diverges or converges then g(x) does the same.. I dont understand why I cant let the smaller one be g(x) and come up with an f(x) that is greater...

Answer 8

that is not true

Answer 9

Zarkon are you talking about his last post or mine?

Answer 10

Misterchyme post

Answer 11

this

f(x)≥g(x)
and f(x) either diverges or converges then g(x) does the same

is not ture

Answer 12

it is possible for f(x) to diverge but g(x) to converge

Answer 13

I thought that was for series not improper integrals

Answer 14

\[g(x)=\frac{1}{x^2}\]
\[f(x)=x^2\]

clearly \[g(x)\leq f(x)\]

\[\int\limits_{1}^{\infty}g(x)dx\leq\infty\]

but

\[\int\limits_{1}^{\infty}f(x)dx=\infty\]

Answer 15

because my book.. in comparison test for improper integrals says..
Suppose that f and g are continious functions with F(x)>g(x)>0 for x>a
If f(x) converges then g(x) converges.. If g(x) diverges then f(x) diverges

Answer 16

that is not what you have above

Answer 17

Exactly. Notice the switch of the order on f(x) and g(x) from converge to diverge.

Answer 18

Its the same concept I was getting at anyways. If f(x) is bigger and converges that FORCES the smaller integral to converge. If g(x) diverges and is a SMALLER integral then obviously the bigger one diverges as well.

Answer 19

Remember, an integral is just an infinite series :P

Answer 20

For some reason I am completely lost

Answer 21

I am prob missing something really simple that is causing this confusion

Answer 22

you know this correct?

if \[g(x)\leq f(x)\]
for all \[x\geq a\]
then

\[\int\limits_{a}^{b}g(x)dx\leq \int\limits_{a}^{b}f(x)dx\]

or to extend it a little

\[\int\limits_{a}^{\infty}g(x)dx\leq \int\limits_{a}^{\infty}f(x)dx\]

Answer 23

the integral preserves the order

Answer 24

Yea i get that, but I dont understand how/what function to use for comparison test.. Something just isnt clicking right now. How do I know wether to chose a function that is less than the original function or greater then is my dilemma

Answer 25

a lot of the time you end up comparing the original function to

\[\frac{1}{x^p}\]

Answer 26

\[\int\limits_{a}^{\infty}\frac{1}{x^p}dx\]

converges for p>1 and diverges otherwise

Answer 27

I should specify that a>0 ;)

Answer 28

well for instance my original function was \[\int\limits_{1}^{\infty} x/(\sqrt(1+x^6)\]

How do I know whether to use f(x)=x which is \[\ge x/(\sqrt(1+x^6)\]
or to use \[1/\sqrt(1+x^6) \le \ x/(\sqrt(1+x^6)\]

Answer 29

\[\frac{x}{\sqrt{1+x^6}}\leq\frac{x}{\sqrt{x^6}}=\frac{x}{x^3}=\frac{1}{x^2}\]

Answer 30

see, that stuff just doesnt pop out to me.. I see it now that you have pointed it out Lol.

Answer 31

do you know now if it will converge or diverge based on what i did?

the integral from 1 to infinity

Answer 32

yes, it will converge because 2>1 correct? ( I hope so!)

Answer 33

yes

Answer 34

Well now that I know what to look for, I should be ok ( hopefully) with a little more practice. Thanks for your help!