Question 1

There is something that always confuses me: the natural logarithm function "grows" less and less for big values of x (as seen at 2:01 on the video)... But why, then, doesn't it have a limit value, since its derivative seems to be zero at infinity?

Accepted Answer

Well ln(x) is the inverse function of e^x. That means that the range of ln(x) is the same as the domain of e^x. Since the domain of e^x is the set of real numbers, that means the range of ln(x) is also the set of real numbers.
Or in other words: you can put anything into e^x => you can get out anything out of ln(x).

Question 2

So, f(x)= x^(-1) has an infinite area from x=1 to infinity, but f(x)=x^(-2) has a FINITE area from x=1 to infinity. 
What I'm wondering is: is there any exponent between -1 and -2 that is the 'divider' between the divergent and the finite area under the curve? That is, x raised to anything less than that exponent is of finite are (between x=1 and infinity), and x raised to anything greater (closer to x^(-1)) is infinite/ divergent.

Accepted Answer

I'm not a math teacher, but from working out a few problems, it seems like the dividing line is x^(-1) itself.

At x^(-1), of course, the integral becomes a natural log, as Sal just solved for above.

If the exponent of x is less than -1, then the integral of the original expression will be some constant multiplied by x^( a negative number). When we evaluate the limit, the lower bound (1) produces some constant, but the other term, when we substitute n for x, now has n in the denominator. As n goes to infinity, this term disappears because 1/infinity is zero (well, not really, but close enough), leaving us with a constant, and thus, a finite solution.

However, if the exponent is greater than -1, when we integrate the original expression we will get a constant times x^(a positive number). When we substitue n for x and n goes to infinity, this term goes to infinity, giving us an infinite solution.

Edited for clarity.

Question 3

Very interesting  1/x^2 has an area of 1
while 1/x has an area of infinity.

I can't seem to grasp this weird thing.

Accepted Answer

Maybe think about it this way: 1/x decreases in proportion with x as x grows, so it's always adding in proportion to the amount it's decreasing by.
1/x^2 decreases in proportion to the SQUARE of x as x grows, so the proportion added decreases as x get larger.

Just to mess with you a little more: 1/n^1.000001 also has a finite area.

Question 4

It doesn't make any sense to me that 1/x^2 would have a finite area, while 1/x would have an infinite area. 1/x^2 is essentially just taking 1/x and stretching it differently. I get that they might not have the same area, but it doesn't seem logical that one would have a finite area and the other wouldn't.

Accepted Answer

This is a natural reaction, I think. Maybe you could convince yourself by studying the behaviour of the series Σ(n→∞) 1/n and the series Σ(n→∞) 1/n² and by understanding the proofs for the fact that the first one diverges and the second one converges. You could do this by calculating  ∫(0→∞) 1/xⁿ. This wil result in two situations: for n ≤ 1 this thing diverges, and for n > 1 this thing converges. And by the way -this will maybe help you in your search- they are called, respectively, the (divergent) harmonic series and the (convergent) hyperharmonic series. Also: http://www.youtube.com/watch?v=aKl7Gwh297c

Question 5

How did 1/x become ln |x|?

Accepted Answer

You mean as opposed to ln x ?

The domain of ln x is x > 0.  You can't take the logarithm of something that is <= 0.

Question 6

Give an example of a convergent improper integral

Accepted Answer

An elementary example would be

`∫ [1, ∞) 1/x² dx,`

where `∫ [1, ∞)` means integral over `[1, ∞)`. More generally,

`∫ [1, ∞) 1/xᵃ dx`

converges whenever `a > 1` and diverges whenever `a ≤ 1`. These integrals are frequently used in practice, especially in the comparison and limit comparison tests for improper integrals.

A more exotic result is

`∫ (-∞, ∞) xsin(x)/(x² + a²) dx = π/eᵃ,`

which holds for all `a > 0`. In particular,

`∫ [0, ∞) xsin(x)/(1 + x²) dx = π/(2e).`

Yet another example are the Fresnel integrals

`∫ [0, ∞) sin(x²) dx = ∫ [0, ∞) cos(x²) dx = (1/4)√(2π).`

Question 7

Hi everyone,

Throughout the course of the episode, I understand how and why integrating the function `1/x` over the interval `x = 1` to `x = infinity` works, but I'm confused as to how this idea agrees with the Comparison Theorem, especially pertaining to one of the principles of the theorem saying that if g(x) (i.e. `1/x`) is divergent from `x = 1` to `x = infinity`, then f(x) (i.e. `1/x^3`) is *also* divergent.

In other words, what causes f(x) to diverge if g(x) diverges?

Thank you all,
Aviel Rodriguez

Accepted Answer

1/x^3 is convergent

Using the theorem can't tell you if 1/x^3 converges or diverges when using 1/x.  1/x^3 < 1/x, but since 1/x doesn't converge, we don't know if 1/x^3 does.  You need to find a function less than the original, and the original also has to converge.

For instance  if you used 1/x^2, since that converges and is also greater than 1/x^3 we can conclude 1/x^3 also converges.

You would need another test to conclude 1/x^2 converges initially though.

Now, if a function f(x) diverges and then another function g(x) > f(x) the comparison test tells us that this larger g(x) also diverges.

Question 8

Why did we use n approaches infinity and not just x approaches to infinity? I mean it just seems like an extra step that is later undone.

Accepted Answer

It's not 𝑥 that approaches infinity, but the upper bound on 𝑥.
That's why Sal uses another variable.

Question 9

I have a question, that is mabye possible related to the one below mine...

Since we are looking for the area under (1/x) from 1 to infinity (can be put into the terms of lim n to inf, integral from 1 to n of 1/x), the rate of change, the derivative of the area is 1/x...

as x approaches infinity, 1/x approaches 0, so why does the area continue to increase to infinity when the rate of change is 0...

Accepted Answer

it is because the rate of change doesn't equal 0, it only approaches 0 in the same way that the function approaches 0

Class 12 math (India)

Course: Class 12 math (India) > Unit 10

Divergent improper integral

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