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## AP®︎ Calculus BC (2017 edition)

### Unit 9: Lesson 6

Improper integrals

# Divergent improper integral

Sometimes the value of an infinite integral is, well, infinite. Created by Sal Khan.

## Want to join the conversation?

• 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. • 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.
• 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. • 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.
• 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. • 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
• • • 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).`
• I very much appreciate Arni Häcki's answer to José Guilherme Licio's question, but even though it gives an illuminating intuition as to why ln(x) doesn't get to a "limit" as x approaches infinity, their answer does not address another point. What exactly is wrong with the following argument?

Since the derivative of ln(x), that is, 1/x, has a limit as x approaches ∞ (and this limit is 0), why can't we say that ln(x) "stops growing" at some point, in essentially the same way we say that "f(x) = 1 - 1/x" stops growing as x approaches ∞? We say f(x) "stops growing" precisely because the limit of 1/x as x approaches ∞ is 0. But using this same argument regarding the derivative of ln(x) doesn't seem to work. What is, precisely, the difference, and why does this difference matter?

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