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Infinite area of f(x)=1/x

  1. Jul 1, 2006 #1
    Let [tex]f(x)=\frac{1}{x} , x \geqq 0[/tex]

    [tex]\therefore \int_{0}^{\infty} f(x) dx = \infty[/tex] (1)

    However the function f(x) has an inverse function of f(x), ie. the function f(x) when m,irrored around y=x is the same shape. Does this imply that when x<1 f(x) approaches [itex]\infty[/itex] as fast as f(x) approaches 0 when x>1?

    If so wouldn't:

    [tex]\int_{0}^{\infty} f(x) dx = 2 \times \int_{1}^{\infty} f(x) dx (2)[/tex]

    I understand mathematically that (1) is correct, but when I think about the areas logically I can't understand how lines that approach the respective axis at what appears to me to be the same rate can have one part (x>1) with a non-infinite area and yet have a part ([itex]0 \leqq x < 1[\itex]) which has an infinite area.
     
    Last edited: Jul 1, 2006
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  3. Jul 1, 2006 #2

    StatusX

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    First, the regions that should have the same area (in an imprecise way, since you can't compare infinite quantities) are the areas seperated by the line y=x, so your integral is slightly off (missing the triangle from (0,0) to (1,0) to (1,1)). Second:

    [tex]\int_1^\infty \frac{1}{x} dx = \ln(x)|_1^\infty = ln(\infty)=\infty[/tex]

    So this does not have a non-infinite area.
     
  4. Jul 2, 2006 #3

    HallsofIvy

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    Yes, that's true. If fact, if f(x)= 1/x and g(x)= f-1(x) then
    f(x)/g(x)= 1 for all x so they go to infinity at exactly the same "rate".

    So you are saying [itex]\infty= 2\times\infty[/itex]? That's true is the extended real numbers in which you can do "arithmetic" on "infinity" but not the regular real numbers.

    ?? What?? Why would you say that for x>1 the area is "non-infinite"?
    [tex]\int_1^\infty \frac{1}{x}dx= \infty[/tex]
    also.
    In fact,
    [tex]\int_a^\infty \frac{1}{x}dx= \lim_{x\rightarrow\infty}ln(x)- ln(a)= \infty[/itex]
    and
    [tex]\int_0^a \frac{1}{x}dx= ln(a)- \lim_{\epsilon\rightarrow 0} ln(\epsilon)= \infty[/tex]
     
  5. Jul 2, 2006 #4

    0rthodontist

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    This is true anyway since as mentioned both sides are infinite, but geometrically the argument would go
    [tex]\int_{0}^{\infty} f(x) dx = 2 \times \int_{1}^{\infty} f(x) dx + 1[/tex]
    because of the unit square that is not included when you integrate from 1 in the x or y direction.
     
  6. Jul 3, 2006 #5
    Well... I think what you're saying is that isn't it the same to sum up 1/x from 1 to infinity as it is to sum up 1/x from 0 to 1. Hmm, I think the main problem here is that when you are summing up terms from 1 to infinty you never encounter infinite numbers, whereas if you were summing from 0 to 1 you would (the infinitely small region dx times the height at 0, infinity--technically undefined I guess, which also might be why). I'm pretty sure that this is the main reason you can't, because you can't add infinity.
     
  7. Jul 3, 2006 #6

    0rthodontist

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    No, what he wants to say is that it's the same to sum 1/x from 1 to infinity as it is to sum 1/x from 0 to 1 and then subtract one. Also, just because the height at 0 is infinite doesn't necessarily mean you can't define an integral- you can define the integral as the limiting case as you get closer to 0. Everything about the geometric argument would be stated the same yet deal with finite values if you let f(x) = 1/x2 instead.
     
    Last edited: Jul 3, 2006
  8. Jul 3, 2006 #7
    Oh, I see what he's saying now, so... what is wrong with his argument?
     
  9. Jul 3, 2006 #8

    HallsofIvy

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    The only thing wrong with the argument is that infinity is not a real number and, even in number systems in which "infinity" is defined, it doesn't satisfy the usual rules of arithmetic.
     
  10. Jul 4, 2006 #9
    I understand that infinity is not a real number.

    I was looking at the rate at which 1/x approaches either axis, but I was thinking of 1/x^2 when working with the integrals (which does have a real definite integral for the lower bound=1 and the upper bound =infinity) so I made a fundamental mistake.

    But the graph of [itex]\frac {1}{x^2}[/itex] is not symmetric about y=x so does not approach the y axis and x axis at the same rate, which clears up my understanding of why [itex]\int_{0}^{1} \frac {1}{x^2} dx \neq \int_{1}^{\infty} \frac {1}{x^2} dx[/itex]. It was just a confusion with the exponential and seems to be such a silly mistake.
     
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