Homework Help: Evaluate or show that is divergent

1. Mar 15, 2013

whatlifeforme

1. The problem statement, all variables and given/known data
evaluate or show that is divergent.

2. Relevant equations
$\displaystyle\int_0^3 {\frac{1}{x-1}dx}$

3. The attempt at a solution

$\displaystyle\int_0^1 {\frac{1}{x-1}dx}$

+

$\displaystyle\int_1^3 {\frac{1}{x-1}dx}$

lim$_{b->1^-} \displaystyle\int_0^b {\frac{1}{x-1}dx}$

+

lim$_{a->1^+} \displaystyle\int_a^3 {\frac{1}{x-1}dx}$

∞ - ln|-1| + ln|2| - ∞ ->> would the two infinities cancel out? is this an indeterminate form?

2. Mar 15, 2013

Zondrina

Hm, was the original integral all together and then you split it or was it two different integrals all together?

3. Mar 15, 2013

whatlifeforme

it was all together but there was a discontinuity at 1 where i had split it up into two.

4. Mar 15, 2013

Dick

∞ - ∞ is undefined. No, you can't cancel them out. If the integral diverges on [0,1) then it diverges on [0,3].

5. Mar 15, 2013

Zondrina

Okay so you originally had $\int_{0}^{3} \frac{1}{x-1} dx$

You can make this a bit easier by comparison :

$\int_{0}^{3} \frac{1}{x-1} dx ≥ \int_{0}^{3} \frac{1}{x} dx = lim_{a→0^+} \int_{a}^{3} \frac{1}{x} dx = lim_{a→0^+} [ ln(x) ]_a^3 = ln(3) - lim_{a→0^+} ln(a) → -∞$

6. Mar 15, 2013

Dick

1/(x-1) is NOT greater than 1/x. That's certainly not true at x=(1/2), for example. And there is nothing wrong with what whatlifeforme was doing. The only error was in the conclusions.

7. Mar 15, 2013

Zondrina

Ohhh my mistake. I think I've been dealing with way too many series recently.

8. Mar 15, 2013

whatlifeforme

what conclusions? do ∞ and -∞ not cancel out?

9. Mar 15, 2013

Dick

Yes, that one. Isn't that what I said?

10. Mar 16, 2013

HallsofIvy

The whole point of writing "a" and "b" is to NOT get ∞! You should have
[tex]ln|a-1|- ln|-1|+ ln|2|- ln|b- 1|= ln(2)+ ln|(a- 1)/(b- 1)|.

No, you cannot cancel the two "infinities". With a and b independent, that last term is not 0.

11. Mar 17, 2013

whatlifeforme

i'm still stuck here.

12. Mar 17, 2013

Dick

Why? You showed the integral from 0 to 1 diverges. That means the whole integral diverges. You were done then.

13. Mar 17, 2013

whatlifeforme

are you sure that is a valid conclusion? so i would simply state the integral diverges on 0 to 1 thus it diverges on 0 to 3?

14. Mar 17, 2013

Dick

Yes, I am sure. Infinities can't be 'cancelled'. If it diverges on 0 to 1 then nothing else can fix that.

15. Mar 17, 2013

whatlifeforme

what if i did a direct comparison test.

1/x-1 > 1/x

$\displaystyle\int_0^3 {\frac{1}{x} dx}$

ln|3| - ln|0|= divergent right?

smaller diverges so large diverges.

16. Mar 17, 2013

Dick

If you want to apply the comparison test you need both functions to be nonnegative on [0,3]. 1/(x-1) isn't. It's not even defined at x=1.

17. Mar 17, 2013

whatlifeforme

update: looks like i did not need to subtract ∞ from ∞ after all.

limit$_{x->0-}$ = -∞
limit$_{x->0+}$ = ∞

Thus, I would have:

limit$_{b->1-} ln|x-1| ^{b}_{0}$ = ∞ - ln|-1|

+

limit$_{a->1+} ln|x-1| ^{a}_{3}$ = ln|2| + ∞

thus, ∞ + ln|2| + ∞.

does this look correct?

18. Mar 17, 2013

Dick

Well, no! Your second limit should be from a to 3, not from 3 to a. Just like you said in your original post. The integral from 0 to 1 gives you an infinitely large negative and the integral from 1 to 3 gives an infinitely large positive. Why are you pushing this so hard and what exactly are you trying to accomplish. As I keep telling you, you had the answer to begin with.

19. Mar 17, 2013

whatlifeforme

i'm just really stressed about getting this right. and taking the first part

$\displaystyle\int_0^1 {\frac{1}{x-1} dx}$ and stating that it diverges, thus,

$\displaystyle\int_0^3 {\frac{1}{x-1} dx}$ diverges.

doesn't seem like a solid way of answering the question.

20. Mar 17, 2013

whatlifeforme

also, ln|0| - ln|0| should equal 0. as they are both the same infinity.

some infinities are larger than others, but in this case they should be the same.

21. Mar 17, 2013

haruspex

No, you cannot argue that way. Ln(0) = 2 ln(0), so ln(0) - ln(0) = ln(0)? Once you start trying to arithmetic with infinity (beyond a few permitted operations) you can get any nonsense result. There are only 'different sizes of infinity' in certain specialised meanings of infinity, such as in cardinality of sets and so-called transfinite arithmetic.

22. Mar 17, 2013

Dick

You are pointlessly stressing yourself over an easy question that you had basically correct in the first post. If you don't believe me that if an integral diverges on part of it's range then it diverges on the whole range, then you can thrash around all you want. I'm getting less and less interested in answering these useless posts.