MHB Improper integrals (Comparison Test)

[renyikouniao]

Use the comparison test to find out whether or not the following improper integral exist(converge)?
integral(upper bound:infinity lower bound:2) 1/(1-x^2) dx

Here's my solution for 3),but I think something went wrong

For all x>=2
0<=-(2-2x)<=-(1-x^2) that means: 0<=-1/(1-x^2)<=-1/(2-2x)

and

integral (upper bound : infinity.lower bound : 2)
-1/(2-2x)=1/2lim(t->infinity)[In(2-2t)-In(-2)]

According to Comparison Test
If ) the integral -1/(2-2x) is convergent,then the integral -1/(1-x^2) is convergent.

But the integral -1/(2-2x) is divergent...so I don't know what to do next...

 

[alyafey22]

$$x^2-1=(x-1)(x+1) \geq \sqrt{x} \cdot x $$

for all $$x \geq 3$$ .

 

[renyikouniao]

$$x^2-1=(x-1)(x+1) \geq \sqrt{x} \cdot x $$

for all $$x \geq 3$$ .

Can I state:if the integral (upper bound: a lower bound: b )is convergent/divergent.
then the negative integral (upper bound: a lower bound: b )is convergent/divergent.For ex:If I prove 1/(x^2-1) is convergent,then I can say 1/(1-x^2) is convergent too.

 

[alyafey22]

Can I state:if the integral (upper bound: a lower bound: b )is convergent/divergent.
then the negative integral (upper bound: a lower bound: b )is convergent/divergent.For ex:If I prove 1/(x^2-1) is convergent,then I can say 1/(1-x^2) is convergent too.

Yes , of course . For an integral to be convergent that means it has a value that is not infinite . So if we have the following

$$\int^b_a f(x) \, dx = C$$ converges then

$$\int^a_b f(x) \, dx = -\int^b_a f(x) \, dx = - C $$

Since the integral has a finite value , it obviously converges because this is what it means to converge .

If you are talking about positive functions then the integral means a finite area under the function on the interval [a,b] . Clearly the area will still be finite if we change the limits to be $$[b,a]$$ but since we are traveling on the opposite side it will be with a negative sign .