# 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

Staff Emeritus
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.