# Homework Help: Improper Integral Comparison Proof

1. Aug 26, 2011

### Benefcb

1. The problem statement, all variables and given/known data

Prove or disprove:
$$b\int_b^∞ f(x) dx ≤ \int_b^∞ xf(x) dx$$

for any b≥0 and f(x)≥0

2. Relevant equations

N/A

3. The attempt at a solution

Ok this question has caused me quite some problems. I have come to the conclusion that this needs to be proven rather than disproven. Integrating f(x) whilst multiplying by x will mean that the resulting function is greater than the original without the x. Since this is an improper integral proof this means that really only the lower bound b is of importance.

My main ideas:

- If f(x) is always greater than or equal to 0 this must imply that it either converges on both sides or diverges on both sides. (This may be wrong)

- What happens when the original function is always greater than 0 yet when it is integrated it diverges. Does this mean that if the other function diverges as well that they are equal?

As you can see I am not really sure how to properly work this proof. There are so many situations that I have a feeling the solution must be something much simpler (This is not intended to be a difficult problem)

Thanks for your help! Hopefully i can contribute to this site as well.

2. Aug 26, 2011

### Staff: Mentor

How would you say this without using "it"? What does it refer to in this sentence?
Since the interval over which integration takes place is [b, $\infty$), won't it be true that bf(x) $\leq$xf(x)?

3. Aug 26, 2011

### Benefcb

Well with it I mean the function f(x). If it converges then it will have a maximum turning point otherwise if it diverges it should have a minimum turning point.

bf(x) $\leq$xf(x) is true yet I simply dont know if that is enough. I feel I can vaguely explain the solution yet I have no way of proving it mathematically.

The most simple functions that work in this situation are: 1/x^n, 1/e^x, 1/e^x^n and so forth yet they all have different ways of integrating. Thus if I want to show that it is always greater when x is included I would need to integrate before comparing.

My only other idea is to use the x inclusive function and compare it to the original and then see by how many factors it is greater since the boundaries are the same.

Surely there must be a concept of integration which atleast helps to prove this?

4. Aug 26, 2011

### Staff: Mentor

This doesn't make sense for a couple of reasons. For the first, in this context we're talking about the improper integrals converging or diverging, not the functions. For the second, how do turning points play any kind of role here?

No, you don't get to pick the function.
What is the "x inclusive function"?
There is: if f(x) <= g(x) for all x in some interval, then ∫f(x)dx <= ∫g(x)dx, where the integration is over that same interval.

5. Aug 26, 2011

### Benefcb

Yes I see what you are saying. Think the last proof could work because we are assuming the function works for an improper integral. This would mean that when taking b as a high value for example would only make the original value smaller (the integrated function needs to go to a real number for infinity to be used thus it must be a reciprocal function and thus the higher b the lower the value). Also when calculated before the integration a value between 0 and 1 would mean a lower value than what x is.

So basically: (I call the functions b-function and x function meaning what they are multiplied by)

Before integration proof: When b is between 1 and infinity the b-function has a lower value since it is a reciprocal

After integration proof: When b is between 0 and 1 it will make the value much smaller than simply using the lower bound since it is determined that integrating with an additional x will cause any result to be higher than without the additional x.

Does my thinking (at least my summary) make some sense or is further evidence/proof required?

Thanks again for all your help by the way I hope im not confusing you too much with my thinking.

6. Aug 26, 2011

### Staff: Mentor

What proof? What do you mean when you say "the function works for an improper integral"?
The only thing that you can reasonably assume about f is that it is continuous on the interval [b, $\infty$). You can't assume anything about it being 1/xn or a reciprocal function or whatever.

Your reasoning in the paragraph above is not likely to convince your instructor.
No. Yes.
Think about f(x), and draw some graphs of continuous functions.
This integral
$$\int_b^{\infty}x \cdot f(x)dx$$
either converges to some finite number or it diverges. I would divide the problem in two cases. What you need to show is that the two integrals in your original post are equal or that the one on the left is less than the one on the right. Here "equals" would include the possibility that both integrals diverge.

7. Aug 26, 2011

### Benefcb

Ok I have gone back and rethought the problem. If the function has an integral this must mean it converges and thus converges to 0. In this case the function will be integrated either with or without the additional x in both cases resulting in a continuous convergent to 0 function. However the additional x means that the area must be greater (due to increased y values). As has been determined when the factor b is large it will cause only a minor change since the integrated value is so small (converges to 0) which is smaller than the change the additional x causes.

In case of a divergent function is it not enough to say that the integral does not exist thus no comparison can be made? This would eliminate divergent functions from the problem.

The only problem I really have is how I can prove that the x will be greater than the constant b. The x will cause the area to go up that is for sure yet it is the factor b I am having trouble with.

In bold are the parts most important in my current argument. I will say this is not for a calc class but for a prob. class so the mathematical concepts are not so strictly enforced

8. Aug 26, 2011

### Staff: Mentor

This is not true. For example, the function f(x) = x is continuous on the entire real line, and hence integrable, but this integral diverges.
$$\int_0^{\infty}x dx$$

This means that $$\lim_{B \to \infty}\int_0^Bx dx~=~\infty$$

On the other hand, this integral converges, but converges to 1, not 0.
$$\int_0^{\infty}\frac{dx}{x^2}$$

This example pretty well wipes out your next paragraph.
Since f(x) >= 0, the only way the integral could diverge (let's not refer to divergent functions) is that the value of the integral is infinity. If both integrals are divergent, they will both be infinite, and we can call them equal.
That's easy. In both integrals the left endpoint of the integration interval is b. x will of course always be >= b.
I don't know about that. If your argument makes no sense, I'm pretty sure you'll get no credit for it.

9. Aug 26, 2011

### Benefcb

Just to make sure you saw in the OP: f(x) >eq 0

Ok so:

- Both diverge to infinity means they are equal
- b-function converges x-function diverges means x-function is always bigger since of course it diverges
- both functions converge means x is greater than the factor b thus the x-function is always bigger.

Now this needs to be put in mathematical terms and hopefully I finally have it somewhat correct.

10. Aug 27, 2011

### Staff: Mentor

Yes, I saw this. That's why I wrote "Since f(x) >= 0, ..." a couple of posts ago.
Please stop saying that the functions converge when what you should be saying is that the integrals converge or diverge.
Use inequalities for the ideas you need to express.
Start talking about the integrands (not b-function/x-function). The first integral could be written as $$\int_b^{\infty} b\cdot f(x)dx$$

11. Aug 27, 2011

### Benefcb

Ill give this another try then:

Case 1: Both integrals diverge thus both limits are infinifty and thus they are equal

$$b\int_b^∞ f(x) dx = \int_b^∞ xf(x) dx$$

Case 2: Both integrals converge.

$$b\int_b^∞ f(x) dx ≤ \int_b^∞ xf(x) dx$$

Would it work to move the b over to the other side and thus have:

$$\int_b^∞ f(x) dx ≤ 1/b\int_b^∞ xf(x) dx$$

Case 2-a: b is greater than 0 less than 1

Since the integrand containing the additional x will be larger than the original integrand multiplying by 1/b will in this case yield an even larger result.

Case 2-b: b is between 1 and infinity:

the additional x will be greater than the 1/b multiple thus it will still remain larger than the original: Thus:

$$b\int_b^∞ f(x) dx ≤ \int_b^∞ xf(x) dx$$

If these statements are correct then all that is left is to explain Case 2-b in greater detail.

12. Aug 27, 2011

### stallionx

Take b , a constant, inside the integral then look for a possible solution. You will need to show b*f<=x*f within the given bounds.

13. Aug 27, 2011

### Benefcb

Im not sure about when to apply b. Would the actual function include b or would it be multiplied by b after integration has been carried out?

If I can leave b out then it is simply a case of showing f(x)<xf(x) which is not so difficult.

This problem is really confusing me.

14. Aug 27, 2011

### stallionx

Integral ( b * f(x) ) <= Integral ( x* f(x) ) from b to infinity where b,f(x)>=0

integrals at (b) , b*f(b)'s cancel and what is left is

Integral ( x* f(x) ) >= 0 from ( b + differential increment, to infinity ) where everything is positive.