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The integral of 1/x

  1. Jan 31, 2007 #1
    if we apply ordinary integration formula to it we'll get [t^0 / 0] regardless of the limits of integration. can someone show me how to prove mathematically that this method is invalid to use and why
     
  2. jcsd
  3. Jan 31, 2007 #2

    mathwonk

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    well, the power rule for antidifferentiation says the antiderivatives of t^n, are of form : constant plus [t^(n+1)]/[n+1], provided n is different from -1.

    so you just have no reason to use that rule on this integral.
     
  4. Jan 31, 2007 #3
    why did they exclude -1?
     
  5. Jan 31, 2007 #4

    radou

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    Your question is equivalent to the question 'Can we divide by zero?'. :wink:
     
  6. Jan 31, 2007 #5

    HallsofIvy

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    What do you mean by "prove mathematically". t^0/0= 1/0 is "undefined". If we set 1/0= x then we have 1= 0*x which is not true for any possible x.

    That's why every text book gives the "power rule" for integration with the proviso "n not equal to -1".

    Many books define a "new function" by
    [tex]log(x)= \int_0^x \frac{1}{t} dt[/tex]
    and then prove that this is, in fact, the usual natural logarithm function.
     
  7. Jan 31, 2007 #6
    if its undefined it mathematically means NO REAL number satisfies it, so why do we go on to find another nway of calculating this integral and why do we use logarithms for it?
     
  8. Jan 31, 2007 #7
    wat i really mean is since 1/0 is UNDEFINED, can u formally assert that even tho its true 1/t still has an integral provided u have known limits?
     
  9. Jan 31, 2007 #8
    help would be much appreciated as this is fundamental in calculus 2 :D pls clear it up 4 me
     
  10. Jan 31, 2007 #9

    HallsofIvy

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    Yes, it is still true that 1/t has an anti-derivative (everywhere except at 0). Since it is continuous for t not equal to 0, it must have an anti-derivative.

    The fact that 1/0 us UNDEFINED simply means that the formula "anti-derivative of [itex]x^n[/itex] is [itex]\frac{1}{n+1}x^{n+1} + C[/itex]" does not apply when n= -1. That doesn't mean that some other formula will not apply.
     
  11. Jan 31, 2007 #10

    mathwonk

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    try thinking of it backwards. let t^n be any power function, where n is any real number at all.

    take the derivative, geting n.t^(n-1).

    can this ever equal 1/t? tht would require n =m 0, but then the zero out fron would still not let this equal 1/t.

    so you never get 1/t this way.

    i.e. the function 1/t is never obtained by differentiating using the power rule.

    so sad.
     
  12. Jan 31, 2007 #11
    alright. now can u show me (prolly by a link) how they established that the natural logarithm function is the 1 that suitsa the integral?
     
  13. Jan 31, 2007 #12

    mathwonk

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    well its a bit abstract. you have to prove somehow that the natural logarithm function equals the area function for1/x, so you need some way to recognize a log function even in disguise.

    so basically you prove that any non constant continuous function L from R+ to R, which saisifies the law L(xy) = Lx)+L(y), must be a log fucntion, and the base of the logs is the unique number a such that L(a) = 1.


    then you prove that since the area function for 1/x (taken from 1 to x) has derivative 1/x, it lso satisifes those laws. hence it must be a log function!


    and by approximating integrals you can show there is some number e between 2 and 3 such that L(e) = 1. and that e is the base.
     
    Last edited: Jan 31, 2007
  14. Jan 31, 2007 #13

    mathwonk

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    lemma: if L'(x) = 1/X, and L(1)=0, THEN L(xy) = L(x) + L(y) for ALL x,y.

    proof: fix y and look at both sides of the equation as functiosn of x. and take their derivatives.

    you get (1/xy)(y) on the left by the chain rule, and 1/x on the right, but these are equal.

    so the fucntions L(xy) and L(x)+L(y) differ by a constant. but pulgging in x=1, gives L(y) = L(y), so in fact these functions are dead equal.

    QED.
     
  15. Feb 1, 2007 #14

    dextercioby

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    No they don't and no they don't. Actually the lower limit in the integral is 1, therefore

    [tex] \ln x=\int_{1}^{x} \frac{1}{t}{}dt [/tex].
     
  16. Feb 1, 2007 #15

    HallsofIvy

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    Ouch!!

    (I shall go hide my head!):blushing:
     
  17. Feb 1, 2007 #16

    mathwonk

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    picky, picky, it was only off by a constant of integration (of + infinity).
     
  18. Feb 1, 2007 #17
    This is interesting I always just asumed it was because [tex]x^{-1} = \frac {1}{0}x^0 =\frac{1}{0}[/tex] or undefined? which of course is nonsense it obviously = [tex] \frac{1}{x}[/tex] or as said a log function, Nice to see there are other reasons.:smile:
     
    Last edited: Feb 1, 2007
  19. Feb 1, 2007 #18

    disregardthat

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    What does the integral mean, does anyone have a link to a proper explanation of it's function?
     
  20. Feb 1, 2007 #19

    arildno

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    Eeh??
    Whatever are you talking about?

    We have defined, for any non-zero, positive x, the function values to be given by a definite integral:
    [tex]\log(x)=\int_{1}^{x}\frac{dt}{t}[/tex]
     
  21. Feb 1, 2007 #20
    It actually means sum, and is a large old fashioned S, essentially a definite integral is simply that:

    [tex] \int_a^b (F)b-(F)a.[/tex]

    It is in a way simillar to summation but different enough to warrant it's own symbol.

     
    Last edited: Feb 1, 2007
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