Proofs with integrals and properties

[mscbuck]

Homework Statement

For a,b > 1 prove that:
$$\int_{1}^{a} (1/t) dt + \int_{1}^{b} (1/t) dt = \int_{1}^{ab} (1/t) dt$$

Homework Equations

Hint: This can be written
$$\int_{1}^{a} (1/t) dt = \int_{b}^{ab} (1/t) dt$$

"Every partition P = {t0, ..., tn} of [1,a] gives rise to a partition P' = {bt0, ..., b(tn)} of [b, ab], and conversely.

The Attempt at a Solution

So far the work that I did was that I replaced the first integral in the original equation with the one from the hint, and I kind of evaluated all of the integrals and found that the left side has a total area of 1. I'm thinking that this should tell me something perhaps about showing that the area can be written in the way of the third integral?

Thanks!

 

[Dick]

The left side doesn't have area 1. I'm thinking you should look at the upper and lower sums of both sides and show they are equal. The width of the intervals in the Riemann sums increases by a factor of b. The value of the function on each interval decreases by a factor of b. Hence?

 

[mscbuck]

Sorry, I must've just evaluated them wrong.

In terms of seeing if their upper and lower sums are equal, how would I go about doing this? I understand that being integrable would imply that we have inf{U(f,P)} = sup{L(f,P)}, but does this become an exercise in notation of various partitions? I'm unsure of how to setup the partitions exactly with the a, b, and ab.

Thanks for your quick help in the right direction though.

 

[Dick]

Sorry, I must've just evaluated them wrong.

In terms of seeing if their upper and lower sums are equal, how would I go about doing this? I understand that being integrable would imply that we have inf{U(f,P)} = sup{L(f,P)}, but does this become an exercise in notation of various partitions? I'm unsure of how to setup the partitions exactly with the a, b, and ab.

Thanks for your quick help in the right direction though.

They basically already gave you the partitions to compare in what I assume is a hint. Take P={t0...tn} of [1,a]. Write out explicitly U(f,P) on [1,a] as a summation. Now take the partition P'={b*t0...b*tn} of [b,ab]. Write out U(f,P') on [b,ab] as a summation. Do you see that they are equal?

 

[mscbuck]

Unfortunately I am not seeing this result. When I am writing my summations, the change in "t" can be replaced by (a-1) and for the second the change in "t" can be replaced by (ab - b)? What would my Mi's be though (i know they are the supremum of f(x) on the interval ti-1 <= t <= ti+1)?

 

[Dick]

Unfortunately I am not seeing this result. When I am writing my summations, the change in "t" can be replaced by (a-1) and for the second the change in "t" can be replaced by (ab - b)? What would my Mi's be though (i know they are the supremum of f(x) on the interval ti-1 <= t <= ti+1)?

It would likely help if you wrote out the summations you are looking at. The change in t of the first summation is (t_{i+1}-t_i). The supremum of f(x) on an interval is at the lower value of t isn't it? 1/x is decreasing.

 

[mscbuck]

For the first:
$$U(f,P) = \sum Mi(a-1)$$

and $$U(f,P') = \sum Mi(ab-b)$$

In both I replaced the change in "t" with the respective intervals.

where Mi = sup{f(t): ti-1 <= x <= ti} in general

 

[Dick]

For the first:
$$U(f,P) = \sum Mi(a-1)$$

and $$U(f,P') = \sum Mi(ab-b)$$

where Mi = sup{f(t): ti-1 <= x <= ti} in general

The sum should be over each interval in the partition, shouldn't it? They don't all have constant width. The widths in the first partition are t_{i+1}-t_i, not a-1.

 

[mscbuck]

Well I decided to take those outside the sum aftewards, figuring that the sum of ti-1, and ti all together would be the interval that it is on. So that should probably edited (my bad, I'm learning Latex and trying to get good at it but I'm forgetting some other things trying to get syntax down) and those terms should be outside the summation.

But even then, if it is t_{i+1}-t_i, what would I do from there?

 

[Dick]

Well I decided to take those outside the sum aftewards, figuring that the sum of ti-1, and ti all together would be the interval that it is on. So that should probably edited (my bad, I'm learning Latex and trying to get good at it but I'm forgetting some other things trying to get syntax down) and those terms should be outside the summation.

But even then, if it is t_{i+1}-t_i, what would I do from there?

So, the sum is (t_{i+1}-t_i)*f(t_i) before you 'simplified' it, right? You can't collect all of the (t_{i+1}-t_i) terms because they are all multiplying different values of f(t_i). Just leave the sum as it is. Write out a similar sum for U(f,P'). That one you can simplify a bit.