Comparison test for basic integral

[PCSL]

integral 1/(1+x^2)dx from 0 to infinity

I decided to compare that to 1/(1+x) (saying 1/(1+x) > (1/(1+x^2)) but this diverges when the original equation converges. Can someone explain why the integral 1/(1+x) was not a proper choice and what the process would be to find a correct comparison.

 

[Mark44]

integral 1/(1+x^2)dx from 0 to infinity

I decided to compare that to 1/(1+x) (saying 1/(1+x) > (1/(1+x^2)) but this diverges when the original equation converges. Can someone explain why the integral 1/(1+x) was not a proper choice and what the process would be to find a correct comparison.

A better choice is 1/x², which is larger than 1/(1 + x²) for large x.

1/(1 + x) was not a good choice if you believed that the original integral was convergent (which I think you did).

When you use comparison for series or integrals, if you believe your series or integral is convergent, you want to find a series or integrand that 1) converges, and 2) is larger than the one you have.

Iif you believe your series or integral is divergent, you want to find a series or integrand that 1) diverges, and 2) is smaller than the one you have.

No other combinations are applicable.

 

[PCSL]

A better choice is 1/x², which is larger than 1/(1 + x²) for large x.

1/(1 + x) was not a good choice if you believed that the original integral was convergent (which I think you did).

When you use comparison for series or integrals, if you believe your series or integral is convergent, you want to find a series or integrand that 1) converges, and 2) is larger than the one you have.

Iif you believe your series or integral is divergent, you want to find a series or integrand that 1) diverges, and 2) is smaller than the one you have.

No other combinations are applicable.

can you explain why 1/x^2 converges? The integral would be -1/x and from 0 to infinity that would be -1/(infinity)+1/0 wouldn't it? I know that 1/x^2 converges because I remember hearing that in class but i don't get why. Thank you very much for your help, I appreciate it a ton.

 

[PCSL]

Also, any general tips on comparison theorum would be awesome. My first test is in three days (calc II) and that is the only idea I don't fully understand.

 

[Mark44]

can you explain why 1/x^2 converges? The integral would be -1/x and from 0 to infinity that would be -1/(infinity)+1/0 wouldn't it? I know that 1/x^2 converges because I remember hearing that in class but i don't get why. Thank you very much for your help, I appreciate it a ton.

No, you can never substitute infinity in for a number as you have done.

In the comparison I was thinking of, the limits are 1 and infinity. It shouldn't matter that the lower limit is different, since all we're doing is ignoring a small (and finite) part of the aream beneath the graph of y = 1/(1 + x²).

To carry out the integral you asked about, you need to use limits.

[tex]\int_1^{\infty}\frac{dx}{x^2} = \lim_{b \to \infty} \left.\frac{-1}{x}\right|_1^b[/tex]
[tex]=\lim_{b \to \infty}-1/b + 1/1 = 1[/tex]
As long as the lower limit of integration is positive,
[tex]\int_a^{\infty}\frac{dx}{x^2}[/tex]
converges. That's why I said that 1/x² was a better choice "for large x."

 

[PCSL]

No, you can never substitute infinity in for a number as you have done.

In the comparison I was thinking of, the limits are 1 and infinity. It shouldn't matter that the lower limit is different, since all we're doing is ignoring a small (and finite) part of the aream beneath the graph of y = 1/(1 + x²).

To carry out the integral you asked about, you need to use limits.

[tex]\int_1^{\infty}\frac{dx}{x^2} = \lim_{b \to \infty} \left.\frac{-1}{x}\right|_1^b[/tex]
[tex]=\lim_{b \to \infty}-1/b + 1/1 = 1[/tex]
As long as the lower limit of integration is positive,
[tex]\int_a^{\infty}\frac{dx}{x^2}[/tex]
converges. That's why I said that 1/x² was a better choice "for large x."

The limit is from 0 to infinity though, that is why I'm confused.

 

[Mark44]

Is the goal here to evaluate the integral or to just show that it is convergent? If all you need to do is show that the integral converges, then what I said earlier is applicable. It doesn't matter that the lower limit of integration is 1 instead of 0. For your function, the important thing is what happens as x gets large.

 

[PCSL]

Is the goal here to evaluate the integral or to just show that it is convergent? If all you need to do is show that the integral converges, then what I said earlier is applicable. It doesn't matter that the lower limit of integration is 1 instead of 0. For your function, the important thing is what happens as x gets large.

Alright. So does 1/0 not mean that the integral diverges? and I am supposed to evaluate the integral which I understand how to do. I really can't see how 1/x^2 from 0 to infinity doesn't diverge. Thanks for bearing with me...

 

[Mark44]

1/0 is meaningless. If all you need to do is evaluate the integral, that's straightforward, but since this is an improper integral (because of the infinite limit of integration), you'll need to use a limit to do that. See post #5 for an example of this technique.

You still seem to be struggling at understanding what I did. The fact that my integral runs from 1 to infinity instead of from 0 to infinity DOESN'T MATTER. As I said before, I am ignoring a small amount of area under the graph of y = 1/(1 + x²), and THAT DOESN'T MATTER, EITHER. The important thing is what happens when x gets very large, AND THAT DOES MATTER.

 

[PCSL]

1/0 is meaningless. If all you need to do is evaluate the integral, that's straightforward, but since this is an improper integral (because of the infinite limit of integration), you'll need to use a limit to do that. See post #5 for an example of this technique.

You still seem to be struggling at understanding what I did. The fact that my integral runs from 1 to infinity instead of from 0 to infinity DOESN'T MATTER. As I said before, I am ignoring a small amount of area under the graph of y = 1/(1 + x²), and THAT DOESN'T MATTER, EITHER. The important thing is what happens when x gets very large, AND THAT DOES MATTER.

I'll talk to my teacher tomorrow. You don't have to reply again, I don't want to annoy you with my inability to understand.

[tex]\int_0^{\infty}\frac{dx}{x^2} = \lim_{c \to \infty} \left.\frac{-1}{x}\right|_0^c

=\lim_{c \to \infty}-1/c + 1/0 = 0 + 1/0[/tex] ?

 

[romsofia]

What mark44 is trying to say, is that if you need to see if the TOTAL area converges or diverges then substituting 0 for 1 in your original function will create a finite area, then we can use that in the comparison instead of making it more complicated by going from 0 to inf when comparing it to 1/x^2.

But like he said, are you just seeing if the integral converges or do you need to actually evaluate it...?

 

[Mark44]

I am not evaluating this integral:

[tex]\int_0^{\infty}\frac{dx}{x^2} = \lim_{c \to \infty} \left.\frac{-1}{x}\right|_0^c

=\lim_{c \to \infty}-1/c + 1/0 = 0 + 1/0[/tex] ?

I am evaluating this integral:

[tex]\int_1^{\infty}\frac{dx}{x^2} = \lim_{c \to \infty} \left.\frac{-1}{x}\right|_1^c

=\lim_{c \to \infty}-1/c + 1/1 = 1[/tex]

 

[PCSL]

My teacher never said we could change the limit for the comparison theorem. That is very helpful - thanks for your help.

and romsofia, the actual limit was not my problem, although we were supposed to evaluate it.