# Basic integration

1. Apr 29, 2016

### Scholar1

1. The problem statement, all variables and given/known data
∫ from 2 to 6 of dx/(x-4)

2. Relevant equations
None

3. 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?

2. Apr 29, 2016

### Math_QED

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

3. Apr 29, 2016

### Scholar1

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

4. Apr 29, 2016

### Math_QED

You integrate from 2 to 6. 4 is in this interval. Your function is not continuous in x = 4.

5. Apr 29, 2016

### Scholar1

So what do I split into 2 integrals?

6. Apr 29, 2016

### Ray Vickson

You are dealing with an improper integral, so you need to define what it means. In your case, you need to evaluate
$$I(a,b) = \int_2^a \frac{dx}{x-4} + \int_b^6 \frac{dx}{x-4},$$
then see of the limit of $I(a,b)$ exists as $a \uparrow 4$ and $b \downarrow 4$.

7. Apr 29, 2016

### Scholar1

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

8. Apr 29, 2016

### Ray Vickson

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 Cauchy Principal Value integral, 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.

9. Apr 29, 2016

### Scholar1

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. Apr 29, 2016

### Ray Vickson

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. Apr 29, 2016

### nfcfox

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.

12. Apr 29, 2016

### vela

Staff Emeritus
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.

13. Apr 29, 2016

### Ray Vickson

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. Apr 29, 2016

### Scholar1

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. Apr 29, 2016

### Ray Vickson

Fair enough.

For the record: the integral (as a general improper integral) does not exist; however, the Cauchy value = 0.

16. Apr 29, 2016

### nfcfox

17. Apr 29, 2016

### SteamKing

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

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