improper integration

[logistic_guy]

here is the question

Solve $\displaystyle \int_{-1}^{1}\frac{1}{x}dx$.


my attemb
$\displaystyle \int_{-1}^{1}\frac{1}{x}dx = \ln x\big|_{-1}^{1} = \ln 1 - \ln -1 = 0 - \ln -1$
if i evaluate $\displaystyle \ln - 1$ in my calculator i get error
what i do wrong?

 

[Dr.Peterson]

here is the question

Solve $\displaystyle \int_{-1}^{1}\frac{1}{x}dx$.


my attempt
$\displaystyle \int_{-1}^{1}\frac{1}{x}dx = \ln x\big|_{-1}^{1} = \ln 1 - \ln -1 = 0 - \ln -1$
if i evaluate $\displaystyle \ln - 1$ in my calculator i get error
what i do wrong?

What did you do wrong? Just about everything!

What does it mean that this is an improper integral?

What is the domain of the logarithm? (The calculator knows, and so should you.)

And what is the integral of 1/x? Look it up, don't just vaguely recall it!

 

[khansaheb]

he domain into two there is the question

Solve $\displaystyle \int_{-1}^{1}\frac{1}{x}dx$.


my attemb
$\displaystyle \int_{-1}^{1}\frac{1}{x}dx = \ln x\big|_{-1}^{1} = \ln 1 - \ln -1 = 0 - \ln -1$
if i evaluate $\displaystyle \ln - 1$ in my calculator i get error
what i do wrong?

(1/x) loses continuity at x= 0.

Thus you have to integrate by splitting domain into two parts.

 

[logistic_guy]

What did you do wrong? Just about everything!

What does it mean that this is an improper integral?

i think it mean we've finite undefined points in the domain of integration

What is the domain of the logarithm? (The calculator knows, and so should you.)

i think $\displaystyle >0$

And what is the integral of 1/x? Look it up, don't just vaguely recall it!

i already do it $\displaystyle \ln x$

(1/x) loses continuity at x= 0.

Thus you have to integrate by splitting domain into two parts.

i feel there's something wrong in my answerhow to split?

 

[topsquark]

i think it mean we've finite undefined points in the domain of integration


i think $\displaystyle >0$


i already do it $\displaystyle \ln x$


i feel there's something wrong in my answerhow to split?

This is basic Calculus I. Look it up in your textbook.

-Dan

 

[Dr.Peterson]

i already do it $\displaystyle \ln x$

Once again, have you looked it up to see if you might be wrong??

Lists of integrals - Wikipedia

en.wikipedia.org

Do you see what you are omitting? can you see why this answer is better than yours?

 

[fresh_42]

We can show you the paths to your problems' solutions, but we cannot understand them for you. Maybe you should solve one question after the other instead of posting dozens of them in very different fields all at once. It requires practice. Quantity isn't really helpful.

 

[logistic_guy]

This is basic Calculus I. Look it up in your textbook.

-Dan

students forget basics, so it's normal if i don't remeber

Once again, have you looked it up to see if you might be wrong??

Lists of integrals - Wikipedia

en.wikipedia.org

Do you see what you are omitting? can you see why this answer is better than yours?

you're talking about $\displaystyle \int \frac{1}{x} dx = \ln x + c$

but the integration in the op is $\displaystyle \int_{a}^{b} \frac{1}{x} dx = \ln x\big|_{a}^{b}$
i don't have to write the $\displaystyle c$ and i think i get this idea from discussing integration in other threads

We can show you the paths to your problems' solutions, but we cannot understand them for you. Maybe you should solve one question after the other instead of posting dozens of them in very different fields all at once. It requires practice. Quantity isn't really helpful.

i like this idea. i'll write it down and i'll think about it

if i'm advance engineering and i survive mechanical and chemical engineering while i'm an artist
do you think a bunch of questions in different fields can overcome me?

don't worry about me, i'm a multitasking student
am i jack of all trades, master of none?

 

[topsquark]

students forget basics, so it's normal if i don't remeber

And I'm telling you that you need to go back and do an intense review. You have forgotten too much.

if i'm advance engineering and i survive mechanical and chemical engineering while i'm an artist
do you think a bunch of questions in different fields can overcome me?

Apparently, yes!

don't worry about me, i'm a multitasking student
am i jack of all trades, master of none?

I think that you try to memorize how to do specific problems, and do not understand the concepts behind the solutions.

-Dan

 

[Dr.Peterson]

you're talking about $\displaystyle \int \frac{1}{x} dx = \ln x + c$

but the integration in the op is $\displaystyle \int_{a}^{b} \frac{1}{x} dx = \ln x\big|_{a}^{b}$
i don't have to write the $\displaystyle c$ and i think i get this idea from discussing integration in other threads

No, You have completely missed the real issue.

Did you even try to look at my link? Here is an image:

​

The point I am making is the absolute value, which will be taught in any textbook. In particular, this is the answer to one of your questions:

$\displaystyle \int_{-1}^{1}\frac{1}{x}dx = \ln x\big|_{-1}^{1} = \ln 1 - \ln -1 = 0 - \ln -1$
if i evaluate $\displaystyle \ln - 1$ in my calculator i get error
what i do wrong?

You get an error because -1 is not in the domain of the log.

And you don't need to evaluate $\ln(-1)$, because it should be $\ln|-1|=\ln(1)=0$.

If you worked on actually studying one subject at a time in order to master it, you wouldn't waste so much time missing everything important.

 

[logistic_guy]

No, You have completely missed the real issue.

Did you even try to look at my link? Here is an image:

View attachment 38965​

The point I am making is the absolute value, which will be taught in any textbook. In particular, this is the answer to one of your questions:

You get an error because -1 is not in the domain of the log.

And you don't need to evaluate $\ln(-1)$, because it should be $\ln|-1|=\ln(1)=0$.

If you worked on actually studying one subject at a time in order to master it, you wouldn't waste so much time missing everything important.

see what happen if i follow what you show me

$\displaystyle \int_{-1}^{1}\frac{1}{x}dx = \ln|x|\bigg|_{-1}^{1} = \ln|1| - \ln|-1| = \ln 1 - \ln 1 = 0 - 0 = 0$

it can't be correct answer. there's something wrong i don't see

khan say i split the integration. i don't know how to do that

 

[fresh_42]

$1/x$ isn't defined at $x=0.$ The function values near zero are plus and minus infinity. Splitting the integral by
$$\int_{-1}^{1} \dfrac{dx}{x}= \int_{-1}^0 \dfrac{dx}{x}+\int_0^{1} \dfrac{dx}{x}$$ does not help. The function cannot be integrated on $[-1,1]$. All you can do is calculate $$\int_{-1}^{-1/n} \dfrac{dx}{x}+\int_{1/n}^{1} \dfrac{dx}{x} .$$ If you approach zero at the same pace from both sides, then the integrals cancel out since the area on the left is minus the area on the right; but only if you approach zero equally from both sides. Otherwise, you could generate any value. The limits $n \to \infty$ do not exist.

 

[logistic_guy]

$1/x$ isn't defined at $x=0.$ The function values near zero are plus and minus infinity. Splitting the integral by
$$\int_{-1}^{1} \dfrac{dx}{x}= \int_{-1}^0 \dfrac{dx}{x}+\int_0^{1} \dfrac{dx}{x}$$ does not help. The function cannot be integrated on $[-1,1]$. All you can do is calculate $$\int_{-1}^{-1/n} \dfrac{dx}{x}+\int_{1/n}^{1} \dfrac{dx}{x} .$$ If you approach zero at the same pace from both sides, then the integrals cancel out since the area on the left is minus the area on the right; but only if you approach zero equally from both sides. Otherwise, you could generate any value. The limits $n \to \infty$ do not exist.

so which statement is more correct

$\displaystyle \int_{-1}^{1}\frac{1}{x}dx = 0$
or
$\displaystyle \int_{-1}^{1}\frac{1}{x}dx =$diverges

 

[Dr.Peterson]

$\displaystyle \int_{-1}^{1}\frac{1}{x}dx = \ln|x|\bigg|_{-1}^{1} = \ln|1| - \ln|-1| = \ln 1 - \ln 1 = 0 - 0 = 0$

it can't be correct answer. there's something wrong i don't see

khan say i split the integration. i don't know how to do that

When you don't know how to do something, you look it up! You don't need to bother other people.

Here's one textbook's explanation; see the last few examples:

Calculus II - Improper Integrals

In this section we will look at integrals with infinite intervals of integration and integrals with discontinuous integrands in this section. Collectively, they are called improper integrals and as we will see they may or may not have a finite (i.e. not infinite) value. Determining if they...

tutorial.math.lamar.edu

 

[logistic_guy]

You don't need to bother other people.

if i bother you or anyone else, it mean i'm a bad boy
i feel i'm a good boy

Here's one textbook's explanation; see the last few examples:

Calculus II - Improper Integrals

In this section we will look at integrals with infinite intervals of integration and integrals with discontinuous integrands in this section. Collectively, they are called improper integrals and as we will see they may or may not have a finite (i.e. not infinite) value. Determining if they...

tutorial.math.lamar.edu

i think i understand exampel $\displaystyle 7 \int_{-2}^{3}\frac{1}{x^3} dx$

it say we stop solving the integration if any part of it diverge

$\displaystyle \int_{-1}^{1}\frac{1}{x} dx = \int_{-1}^{0}\frac{1}{x} dx + \int_{0}^{1}\frac{1}{x} dx = \lim_{a \to 0^{-}}\int_{-1}^{a}\frac{1}{x} dx + \lim_{a \to 0^{+}} \int_{a}^{1}\frac{1}{x} dx$

let me check the left side

$\displaystyle \lim_{a \to 0^{-}}\int_{-1}^{a}\frac{1}{x} dx = \lim_{a \to 0^{-}}\ln|x|\bigg|_{-1}^{a} = \lim_{a \to 0^{-}}\ln|a| - \ln|-1| = \lim_{a \to 0^{-}}\ln |a|$

i need to see graph of this



this give me idea

$\displaystyle \lim_{a \to 0^{-}}\ln |a| = -\infty$

it say i stop at this point and it mean the integration divergent. i'm not sure if this the correct way to do it