- #1

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## Homework Statement

∫ from 2 to 6 of dx/(x-4)

## Homework Equations

None

## The Attempt at a Solution

u=x-4

du=dx

ln abs value x-4 abs value

ln2-ln2=0

What am I doing wrong?

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- Thread starter Scholar1
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- #1

- 39

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∫ from 2 to 6 of dx/(x-4)

None

u=x-4

du=dx

ln abs value x-4 abs value

ln2-ln2=0

What am I doing wrong?

- #2

member 587159

How does the graph of 1/(x-4) look like ? What do you know when x = 4?

- #3

- 39

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How does the graph of 1/(x-4) look like ? What do you know when x = 4?

Vertical asymptote what does that have to do with this integration?

- #4

member 587159

You integrate from 2 to 6. 4 is in this interval. Your function is not continuous in x = 4.Vertical asymptote what does that have to do with this integration?

- #5

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So what do I split into 2 integrals?

- #6

Ray Vickson

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## Homework Statement

∫ from 2 to 6 of dx/(x-4)

## Homework Equations

None

## The Attempt at a Solution

u=x-4

du=dx

ln abs value x-4 abs value

ln2-ln2=0

What am I doing wrong?

You are dealing with an improper integral, so you need to

[tex] I(a,b) = \int_2^a \frac{dx}{x-4} + \int_b^6 \frac{dx}{x-4}, [/tex]

then see of the limit of ##I(a,b)## exists as ##a \uparrow 4## and ##b \downarrow 4##.

- #7

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So since ln4-4=0 its divergent regardless of all the other stuff?

- #8

Ray Vickson

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So since ln4-4=0 its divergent regardless of all the other stuff?

I cannot tell who you are responding to, but if it is to my post #6, then you need to check if the limit of ##I(a,b)## exists.

There is also the issue of whether or not you mean the

- #9

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I cannot tell who you are responding to, but if it is to my post #6, then you need to check if the limit of ##I(a,b)## exists.

There is also the issue of whether or not you mean theCauchy Principal Valueintegral, which can exist even if a more general type does not. The Cauchy principal value (if it exists) would be ##\lim_{\epsilon \downarrow 0} I(4-\epsilon, 4 + \epsilon)## in the notation of post #6.

I apologize I was replying to you. I am only Calc 2 so I don't know what the Cauchy Principal Value is. I am confused how to check if the limit exist. Please show me.

- #10

Ray Vickson

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I apologize I was replying to you. I am only Calc 2 so I don't know what the Cauchy Principal Value is. I am confused how to check if the limit exist. Please show me.

I have just explained what the Cauchy Principal value is, by giving an actual formula.

PF rules do not allow me to complete your problem for you.

- #11

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He just said he doesn't know what it is, you really think he'd understand that equation? I'm in AP Calc BC which is the calc 2 equivalent and I have never seen anything similar to that equation. You should explain it to him from a calculus 2 standpoint.I have just explained what the Cauchy Principal value is, by giving an actual formula.

PF rules do not allow me to complete your problem for you.

- #12

vela

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For your purposes, the integral doesn't exist for the reason Math_QED gave back in post #4. You can't just mechanically follow a procedure and get an answer. You have to know the limitations.So since ln4-4=0 its divergent regardless of all the other stuff?

- #13

Ray Vickson

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He just said he doesn't know what it is, you really think he'd understand that equation? I'm in AP Calc BC which is the calc 2 equivalent and I have never seen anything similar to that equation. You should explain it to him from a calculus 2 standpoint.

Yes, I would hope that he would understand that equation after working at it for a while---not instantly, maybe, but after spending some time on it. Alternatively, he could Google 'Cauchy principal value' and find numerous web pages dealing with the topic, although skipping the Mathworld and Wikipedia pages on it is probably a good idea for a beginning student.

Anyway, I am assuming that 'Scholar1' (as a Calc 2 student) is capable of evaluating both of the very elementary integrals in the definition of ##I(a,b)##, and that he can then put ##a = 4 - \epsilon, b = 4 + \epsilon## to see what happens. Am I really expecting too much?

By the way: one reason that he/she might be unfamiliar with the Cauchy Principal value is that it is usually found only in some applications in Physics, Engineering and the like. Some writers regard it as a bit of meaningless nonsense in general.

- #14

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Yes, I would hope that he would understand that equation after working at it for a while---not instantly, maybe, but after spending some time on it. Alternatively, he could Google 'Cauchy principal value' and find numerous web pages dealing with the topic, although skipping the Mathworld and Wikipedia pages on it is probably a good idea for a beginning student.

Anyway, I am assuming that 'Scholar1' (as a Calc 2 student) is capable of evaluating both of the very elementary integrals in the definition of ##I(a,b)##, and that he can then put ##a = 4 - \epsilon, b = 4 + \epsilon## to see what happens. Am I really expecting too much?

By the way: one reason that he/she might be unfamiliar with the Cauchy Principal value is that it is usually found only in some applications in Physics, Engineering and the like. Some writers regard it as a bit of meaningless nonsense in general.

I am actually in AP Calculus BC I just said calc 2 because people are more familiar with that. It wouldn't serve me well to spend time learning something so specific when I am preparing for an exam in a limited spectrum course.

- #15

Ray Vickson

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I am actually in AP Calculus BC I just said calc 2 because people are more familiar with that. It wouldn't serve me well to spend time learning something so specific when I am preparing for an exam in a limited spectrum course.

Fair enough.

For the record: the integral (as a general improper integral)

- #16

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Why are you worried about this integral man? I'm worried about sequences and seriesI am actually in AP Calculus BC I just said calc 2 because people are more familiar with that. It wouldn't serve me well to spend time learning something so specific when I am preparing for an exam in a limited spectrum course.

- #17

SteamKing

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You are dealing with an improper integral, so you need todefinewhat it means. In your case, you need to evaluate

[tex] I(a,b) = \int_2^a \frac{dx}{x-4} + \int_b^6 \frac{dx}{x-4}, [/tex]

then see of the limit of ##I(a,b)## exists as ##a \uparrow 4## and ##b \downarrow 4##.

I think the more conventional form of writing the Cauchy PV integral is:

$$I(a,b)=\lim_{ε \rightarrow0+} \left [ \int_2^{4-ε} \frac{dx}{x-4} + \int_{4+ε}^6 \frac{dx}{x-4} \right ]$$

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