Evaluate or show that is divergent

[whatlifeforme]

Homework Statement

evaluate or show that is divergent.

Homework Equations

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

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?

 

[STEMucator]

Homework Statement

evaluate or show that is divergent.

Homework Equations

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

The Attempt at a Solution

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

+


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


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

+

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


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

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

 

[whatlifeforme]

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

 

[Dick]

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

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

 

[STEMucator]

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

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

You can make this a bit easier by comparison :

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

 

[Dick]

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

You can make this a bit easier by comparison :

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

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.

 

[STEMucator]

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.

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

 

[whatlifeforme]

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.

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

 

[Dick]

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

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

 

[HallsofIvy]

Homework Statement

evaluate or show that is divergent.

Homework Equations

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

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?

The whole point of writing "a" and "b" is to NOT get ∞! You should have
$$ln|a-1|- ln|-1|+ ln|2|- ln|b- 1|= ln(2)+ ln|(a- 1)/(b- 1)|.<br /> <br /> No, you cannot cancel the two "infinities". With a and b independent, that last term is not 0.$$

 

[whatlifeforme]

i'm still stuck here.

 

[Dick]

i'm still stuck here.

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

 

[whatlifeforme]

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

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?

 

[Dick]

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?

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

 

[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.

 

[Dick]

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.

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.

 

[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?

 

[Dick]

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?

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.

 

[whatlifeforme]

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.

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.

 

[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.

 

[haruspex]

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.

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.

 

[Dick]

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.

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.