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Divergent improper integral

AP.CALC:
LIM‑6 (EU)
,
LIM‑6.A (LO)
,
LIM‑6.A.1 (EK)
,
LIM‑6.A.2 (EK)
Sometimes the value of an infinite integral is, well, infinite. Created by Sal Khan.

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  • blobby green style avatar for user José Guilherme Licio
    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?
    (31 votes)
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    • piceratops ultimate style avatar for user Arni Häcki
      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).
      (88 votes)
  • leaf red style avatar for user NP
    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.
    (24 votes)
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    • mr pants teal style avatar for user Finrod Felagund
      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.
      (21 votes)
  • male robot hal style avatar for user Eldi
    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.
    (8 votes)
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    • aqualine ultimate style avatar for user Andrew
      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.
      (15 votes)
  • blobby green style avatar for user Bryan Ferris
    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.
    (6 votes)
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    • aqualine seed style avatar for user Zeta
      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
      (7 votes)
  • piceratops ultimate style avatar for user Kartikeye
    How did 1/x become ln |x|?
    (4 votes)
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  • blobby green style avatar for user James Rinnovatore
    Give an example of a convergent improper integral
    (2 votes)
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    • leaf grey style avatar for user Qeeko
      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).
      (6 votes)
  • duskpin ultimate style avatar for user Diego Oliveira
    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!
    (4 votes)
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  • starky seed style avatar for user Aviel Rodriguez
    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
    (2 votes)
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    • female robot grace style avatar for user loumast17
      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.
      (2 votes)
  • female robot grace style avatar for user Osama Shammout
    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.
    (2 votes)
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  • leaf green style avatar for user Ashray Manepalli
    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...
    (2 votes)
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Video transcript

Right here I have graphed part of the graph of y is equal to 1/x. And what I'm curious about is the area under this curve and above the x-axis between x equals 1 and infinity. So I want to figure out this area right over here. So let's try to do it. So we could set this up as an improper integral going from 1 to infinity of 1/x dx. Well once again-- actually, let me do that same yellow color. I like that more-- we can view this as the limit as n approaches infinity of the integral from 1 to n of 1/x dx, which we can write as the limit as n approaches infinity of the antiderivative of 1/x, which is the natural log of the absolute value of x. So this is going to be the natural log of the absolute value of x. And the absolute value of x won't really matter so much. We could just say x because we're dealing with positive values of x. But I'll just write it down as the natural log of the absolute value of x between x is 1 and x is n. And so this is going to be equal to the limit as n approaches infinity of-- you evaluate this at n, so you're going to get the natural log-- I could write the absolute value of n, but we know that n is going to be positive. So we could just write the natural log of n minus the natural log of the absolute value of 1, or the natural log of 1. Natural log of 1 is just a 0. e to the 0-th power is equal to 1. So this boils down to the limit as n approaches infinity of the natural log of n. Now this is interesting. Natural log function just keeps getting larger and larger and larger. The natural log function keeps growing and growing and growing like this, albeit at a slower and slower pace, but it keeps growing. The limit as n approaches infinity of the natural log of n is just equal to infinity. So here we do not have a finite area. This is an infinite. This is an infinite area. It's interesting. When this function decreased faster-- when it was 1 over x squared-- we had a finite area. Now we have an infinite area. And so we would say that this integral right over here, this improper integral, is divergent. And we're done.