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Integral of 1/x

  1. Jan 8, 2013 #1
    The integral of 1/x is ln(x). Where does that come from? That always puzzled me. We can continue to take derivatives through x^0 and into the negative integers, and just use the plain old power rule to get the answers. We can do the same for the integral of x all the way from negative exponents through positive exponents with the exception of x^-1. If we try to take the integral here, we get x^0/0, which is 1/0, and is undefined. OK, I get that, but how do we get a natural logarithm out of this undefined expression?
     
  2. jcsd
  3. Jan 8, 2013 #2
    First use the limit definition of a derivative on ln(x) to show that dln(x)/dx=1/x then use the fundamental theorem of calculus to show that the antiderivative of 1/x is ln(x).
     
  4. Jan 8, 2013 #3

    tiny-tim

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    Hi DiracPool! :smile:
    Let's find [itex]\lim_{n\rightarrow 0}\int_1^{e^a} x^{n-1} dx[/itex]

    that's limn->0 (ena - 1)/n

    = limn->0 (na + n2a2/2! + …)/n

    = a = [ln(ea) - ln(1)] :wink:
     
  5. Jan 9, 2013 #4
    Agreed. Using the rule for other exponents here gives us an undefined integral. Does this mean the integral of 1/x is undefined? No, it just means we have to find another way to evaluate the integral (FTC, as bp_psy pointed out,) as there's obviously an expression for the area under the curve 1/x, so long as both bounds are on the same side of the y-axis.

    Just for a few other examples where applying a rule to find a defined expression results in an undefined answer:

    Quotient rule on [itex]\lim\limits_{x\to0}\left(\dfrac xx\right)[/itex]

    L'Hôpital twice on [itex]\lim\limits_{x\to0}\left(\dfrac{x\cdot\sin \left(x\right)}{x^2}\right)[/itex]

    Both of these result in meaningless answers, but both limits are clearly defined. So we have to find another method to solve them.
     
  6. Jan 10, 2013 #5
    Great, thanks for the insight everyone.
     
  7. Jan 10, 2013 #6

    HallsofIvy

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    Many newer Calculus texts handle this by noting, first, that the rule "[itex]\int x^n dx= 1/(n+1) x^{n+1}+ C[/itex]" cannot be used for n= -1 because it would involve dividing by 0 and then defining [itex]ln(x)= \int_1^t dt/t[/itex]
    From that definition one can derive all of the properties of ln(x), including the facts that ln(ab)= ln(a)+ ln(b), that [itex]ln(a^b)= bln(a)[/itex], that ln(x) is invertible, and that its inverse is [itex]e^x[/itex] where is the unique number satisfying 1= ln(e).
     
  8. Jan 16, 2013 #7
    The natural logarithm function [itex]\ln{x}[/itex] has two equivalent definitions.
    1) [itex]\ln{x}[/itex] is the function such that [itex]\exp(\ln{x})=\ln(e^x)=x[/itex], and
    2) [itex]\ln{x}=\int_1^x \frac{dt}{t}[/itex].

    Proof. From definition 1 to 2:
    The derivative of [itex]\exp(\ln{x})[/itex] is [itex]\exp(\ln{x})\frac{d}{dx}\ln{x}[/itex]. But [itex]\exp(\ln{x})=x[/itex], so [itex]x\frac{d}{dx}\ln{x}=\frac{d}{dx}x=1\implies\frac{d}{dx}\ln{x}= \frac{1}{x}[/itex]. By the Fundamental Theorem of Calculus, [itex]\frac{d}{dx}\int_1^x \frac{dt}{t}=\frac{1}{x}[/itex]. Since [itex]f'(x)=g'(x)\implies f(x)=g(x)+C[/itex] for some [itex]C\in\mathbb{R}[/itex], we have [itex]\ln{x}=\int_1^x \frac{dt}{t}[/itex] (as it turns out, [itex]C=0[/itex]).

    From definition 2 to 1:
    Use the substitution [itex]t=e^u,\,\,dt=e^udu[/itex] to get the integral [itex]\ln{x}=\int_{g(1)}^{g(x)} \frac{e^udu}{e^u}[/itex] where [itex]g(e^x)=e^{g(x)}=x[/itex]. From this we can surmise that [itex]g(1)=0[/itex]. The integral simplifies to [itex]\ln{x}=\int_0^{g(x)}du=\left. u\right|_0^{g(x)}=g(x)[/itex]. So [itex]\ln{x}[/itex] satisfies [itex]\exp(\ln{x})=\ln(e^x)=x[/itex].
     
    Last edited: Jan 16, 2013
  9. Jan 17, 2013 #8

    HallsofIvy

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    Since you post that, it is worth noting:
    If we define [itex]ln(x)= \int_1^x dt/t[/itex] it is easy to show that, since 1/t is continuous for all t except 0 ln(x) is defined for all positive x. Of course, 1/x, the derivative of ln(x) (by the fundamental theorem of Calculus) is positive for all positive x, ln(x) is an increasing function which means that ln(x) is a "one-to-one" function mapping the set of all positive numbers to the set or all real numbers. That tells us that it has an inverse and we can define "exp(x)" to be that inverse.

    Now, the crucial point is this: if y= exp(x) then, since exp(x) is the inverse function to ln(x), x= ln(y). If x is not 0, that is equivalent to [itex]1= (1/x)ln(y)= ln(y^{1/x})[/itex]. Going back to the exponential form, [itex]exp(1)= y^{1/x}[/itex] so that [itex]y= exp(x)= (exp(1))^x[/itex]. That is, that inverse function really is just a number (and we can define e= exp(1)) to the x power.
     
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