MHB 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.

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

so:

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

Letting $$u = x^2 + 1$$
and $$du = 2x \, dx$$

after updating the limits

I come up with

$$\int ^{a^2 + 1}_2 u^{-3} \, du$$

I guess my question is: I know that the upper limit is really just infinity$$^2$$ + 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 $$\frac{number}{\infty} = 0$$ but does $$\frac{\infty}{number} = \infty ?$$ (Again, sorry if these sound like ignorant questions, I'm just a bit confused.)

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

$$\therefore$$ converges because $$\frac{1}{8} > 1$$

 

[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 $$\int_{t>0}^\infty \frac{1}{x^a} dx$$

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)

$$\therefore$$ converges because $$\frac{1}{8} > 1$$

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: $$\frac{1}{-2u^2} | ^{a^2 + 1}_2 = 0 + \frac{1}{8} $$

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, $$\therefore$$ 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: $$\frac{1}{-2u^2} | ^{a^2 + 1}_2 = 0 + \frac{1}{8} $$

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, $$\therefore$$ 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!