Evaluate the Improper Integral (III)

[shamieh]

Need someone to check my work, as well as answer a few questions I'm confused about as well.

\(\displaystyle \int ^{\infty}_1 \frac{2x}{(x^2 + 1)^3} \, dx\)

so:

\(\displaystyle lim_{a\to\infty} \int^a_1 \frac{2x}{(x^2 + 1)^3} \, dx\)

Letting \(\displaystyle u = x^2 + 1\)
and \(\displaystyle du = 2x \, dx\)

after updating the limits

I come up with

\(\displaystyle \int ^{a^2 + 1}_2 u^{-3} \, du\)

I guess my question is: I know that the upper limit is really just infinity\(\displaystyle ^2\) + 1 right? So do I really need to worry about writing that substitution in the upper limit or can I just leave it as a

Also I know that \(\displaystyle \frac{number}{\infty} = 0\) but does \(\displaystyle \frac{\infty}{number} = \infty ?\) (Again, sorry if these sound like ignorant questions, I'm just a bit confused.)

I came up with: \(\displaystyle \frac{1}{-2u^2} |^a_2 = [0] + [\frac{1}{8}] \)

\(\displaystyle \therefore\) converges because \(\displaystyle \frac{1}{8} > 1\)

 

[alyafey22]

Re: Evaluate the Improper Integrl (III)

I would expect the integral to converges because the denominator goes faster than the numerator by huge amount. Natural we want to compare the integral to \(\displaystyle \int_{t>0}^\infty \frac{1}{x^a} dx\)

The integral converges if $a>1$.

 

[shamieh]

Re: Evaluate the Improper Integrl (III)

But in the first integral isn't it = 0 so it diverges because a < 1 ?

Therefore if one diverges don't they both?

 

[alyafey22]

Re: Evaluate the Improper Integrl (III)

\(\displaystyle \therefore\) converges because \(\displaystyle \frac{1}{8} > 1\)

We only need the final result to be finite for the integral to converge.

 

[shamieh]

Re: Evaluate the Improper Integrl (III)

Oh I see. I was analyzing it wrong. So it in fact does converge then.

 

[shamieh]

Re: Evaluate the Improper Integrl (III)

We only need the final result to be finite for the integral to converge.

So could I say that: \(\displaystyle \frac{1}{-2u^2} | ^{a^2 + 1}_2 = 0 + \frac{1}{8} \)

since a < 1 we know that it converges.

or

Since we have some finite limit that does exist, we can say that it definitely converges considering that 1/8 = .025 which is < 1

or

Do I not need to worry about the actual value that the limit is itself, \(\displaystyle \therefore\) if it exists we know that it converges?

Sorry, just trying to clarify

 

[Prove It]

Re: Evaluate the Improper Integrl (III)

So could I say that: \(\displaystyle \frac{1}{-2u^2} | ^{a^2 + 1}_2 = 0 + \frac{1}{8} \)

since a < 1 we know that it converges.

or

Since we have some finite limit that does exist, we can say that it definitely converges considering that 1/8 = .025 which is < 1

or

Do I not need to worry about the actual value that the limit is itself, \(\displaystyle \therefore\) if it exists we know that it converges?

Sorry, just trying to clarify

If the integral is a finite number, as it is in this case, then the integral is convergent. The limit being less than 1 is irrelevant, in fact, the integral can equal ANY finite number (greater than 1 as well) and still converge.

 

[shamieh]

Re: Evaluate the Improper Integrl (III)

If the integral is a finite number, as it is in this case, then the integral is convergent. The limit being less than 1 is irrelevant, in fact, the integral can equal ANY finite number (greater than 1 as well) and still converge.

Awesome, thanks so much! Was so confused on this. Now I get it!