Quick question about integral of (1/x)

  1. I know that the integral of 1/x is ln(x)

    but if you follow the conventional method of integrating you raise the power and multiply by the new power, and you get an answer that is obviously wrong

    Would someone mind explaining to me why you cannot integrate this function by normal means?



    Thanks,
    Oscar
     
  2. jcsd
  3. One way to think of it is the following: If we could find the antiderivative by "normal" means, then we would get
    [itex]\int \frac{1}{x}~dx = x^{0} = \textrm{constant}[/itex]
    which is clearly not true from looking at the graph of 1/x

    Another way to look at it: the reason we can do the reverse power rule usually for two reasons: 1) the fundamental theorem of calculus, and 2) the power rule. In other words, it usually works because we know if we find the function F(x) such that [itex]x^{n}[/itex] is the derivative, then we know the integral is equal to F(x). But the power rule only works when n is not zero, since D[[itex]x^{0}[/itex]] = D[1] = 0.
     
  4. lurflurf

    lurflurf 2,327
    Homework Helper

    This is silly.
    You are maybe thinking of
    Integral[x^a]=x^(a+1)/(a+1) (* a!=-1)
    This cannot hold when a=-1 because it would require division by zero.
    We can however use limits to extend the result.
    Integral[x^a]=lim_{a->a} x^(a+1)/(a+1)
    in particular
    Integral[x^-1]=lim_{a->-1} x^(a+1)/(a+1)=log(x)
     
    Last edited: Mar 20, 2009
  5. lanedance

    lanedance 3,307
    Homework Helper

    hmmm.. do you mean

    [tex] \int x^a = ^{lim}_{b\rightarrow a} \frac{x^{b+1}}{b+1} [/tex]
    then
    [tex] \int \frac{1}{x}= ^{lim}_{b\rightarrow -1} \frac{x^{b+1}}{b+1} =ln(x) [/tex]
     
  6. HallsofIvy

    HallsofIvy 40,967
    Staff Emeritus
    Science Advisor

    It is not clear what you are asking. The obvious answer is what lurflurf said. You cannot use the formula
    [tex]\int x^n dx= \frac{1}{n+1} x^{n+1}+ C[/tex]
    when n= -1 because then you would be dividing by 0. I'm not sure why you think of that as "normal means".
     
  7. Think of it this way, the antiderivative of 1/x is the function whose inverse is exactly equal to its own derivative. Let y(x) be the antiderivate of 1/x. Then we have:

    [tex]

    \frac{dy}{dx} = \frac{1}{x}
    [/tex]

    Inverting the Liebniz notation in the way that he intended yields:

    [tex]
    \frac{dx}{dy} = x
    [/tex]

    The last equation says that x' = x, i.e. the function x(y) is equal to its own derivative. This means that x(y) cannot be a polynomial or rational function, since all of those functions change when you differentiate them. It is this perfect property, that the antiderivative of 1/x is the inverse function of the function who is exactly equal to its own derivative, that puts in a class of its own, a special case.
     
  8. Hi,

    Sorry for being unclear.

    My question was supposed to ask why you cannot do the following:

    [tex]\int x^n dx= \frac{1}{n+1} x^{n+1}+ C[/tex]

    for the special case of x-1, I realise that it cannot be done from looking at the graph of y=[tex]1/x[/tex] and because the fraction requiring division by 0 cannot be defined. my query was as to why you could not do this for x-1 (which confinement answered).

    Once again appologies for the lack of clarity in my question, all your responses have cleared some of the clouds from my head :)

    thanks very much,

    Oscar
     
  9. lurflurf

    lurflurf 2,327
    Homework Helper

    Yes, though lim a->a is valid it is confusing to some, I also typoed a into x. Also omiting dx is valid, but confusing to some. Most people denote their favorite base as log, what then is you favorite base if not e?
     
  10. lanedance

    lanedance 3,307
    Homework Helper

    yeah its got to be e, though i like the ln just to avoid any confusion
     
    Last edited: Mar 21, 2009
  11. Greetings:

    d/dx [ln(x)] = lim(h-->0) [(ln(x+h) - ln(x)) / h]

    = lim(h-->0) [(1/h)*ln[(x+h)/x]

    = (1/x) lim(h-->0) [(x/h) ln(1+h/x)]

    = (1/x) lim(h-->0) [ln(1+h/x)^(x/h)]

    = (1/x) ln [lim(h-->0) [(1+(h/x))^(x/h)]] {from lim[f(g(x))] = f(lim[g(x)])}

    = (1/x) ln e

    = 1/x.

    Since d/dx [ln(x)] = 1/x, it follows that intgrl[1/x] = ln x.

    Regards,

    Rich B.
    rmath4u2@aol.com
     
  12. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    You might prefer the following argument for why the anti-derivative should be ln(x) plus some constant C:

    1. We define the arbitrary (non-natural) power of some (positive) variable as follows:
    [tex]x^{a}=e^{a\ln(x)}[/tex]

    2. In terms of natural powers, we define the exponential function itself as follows:
    [tex]e^{y}=1+\sum_{n=1}^{\infty}\frac{y^{n}}{n!}[/tex]

    3. Let us utilize the following definite integral to prove our point:
    We have:
    [tex]\int_{a}^{b}x^{-1+\epsilon}dx=\frac{1}{\epsilon}(b^{\epsilon}-a^{\epsilon})[/tex],
    where a,b,[itex]\epsilon[/itex] are all considered greater than 0.

    4. Now, let us utilize definitions from 1, 2 upon the expression in 3. We gain:
    [tex]\frac{1}{\epsilon}(b^{\epsilon}-a^{\epsilon})=\sum_{n=1}^{\infty}\frac{\epsilon^{n-1}((\ln(b))^{n}-(\ln(a))^{n}}{n!})[/tex]

    5. Thus, in the limit [itex]\epsilon\to{0}[/itex], all but the first double-term of this sum goes to zero, and we retain:
    [tex]\lim_{\epsilon\to{0}}\int_{a}^{b}x^{-1+\epsilon}dx=ln(b)-ln(a)[/tex]
    This is in accordance with what we should have!

    6. Note that this argument shows that the definite integral of some power of x can be defined as a continuous function of the power variable, even in the "special case" where the power variable has the value -1.
     
    Last edited: Mar 21, 2009
  13. I'm new to calculus, but my impression is that an integral should be a function that represents the area under the curve of the function being integrated, which f(x)=ln(x) does not do for f(x)=1/x

    In the graph of 1/x you can see that from 0 toward infinity the area under the curve would come on quickly and almost stop increasing.

    Can anyone shed light on what I'm missing.

    I apologize if this is elementary. I took college algebra 10 years ago, recently took a precalc class at a community college and just finished reading a text book called Brief Calculus.

    Love math and physics by the way and am glad I found this site.
     
  14. To 2^Oscar and the rest of this thread's users:



    --------> [itex]\int{\dfrac{1}{x}}[/itex]dx[itex] \ = \ \ln|x| \ + \ C[/itex]



    Other than missing the arbitrary constant, "C," there must
    be absolute value bars around the "x."



    [itex]*** Edit ***[/itex]
    [itex]\text{Good catch by micromass concerning the missing dx term.}[/itex]
     
    Last edited: Dec 21, 2011
  15. lavinia

    lavinia 2,124
    Science Advisor

    While the answers already given are correct here is a way of looking at it that has always mystified me.

    1/x has the remarkable property that the area under it from 1 to some number is the sum of the areas from 1 to any two numbers whose product equals the number. So if x = yz then the area from 1 to x is the sum of the areas from 1 to y and 1 to z. It is easy to show this using the integral definition of area though I would love to see an elementary proof not using calculus.

    It is not hard to convince yourself that a rational function (quotient of two polynomials) can not have this property. So the logarithm is something new. It's derivative is a rational function but it is not. It is like a portal from the rational to the transcendental.

    A function that compares two laws of addition and/or multiplication is called a homomorphism. The logarithm is a homomorphism from the positive numbers under multiplication to all of the numbers under addition. The exponential function is the reverse homomorphism taking the numbers under addition to the positive numbers under multiplication.
     
    Last edited: Dec 21, 2011

  16. [itex]\text{Not those answers as they concern where the}[/itex]
    [itex]\text{absolute value symbol is missing, not to mention the "+ C."}[/itex]


    Please see the post above yours regarding this.
     
  17. micromass

    micromass 19,689
    Staff Emeritus
    Science Advisor
    Education Advisor

    You forgot the dx in your post above. So your answer is also incorrect.
     
  18. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    Somehow (to me at least) related to lavinia's post is the interesting fact discovered by Euler

    [tex] \mbox{for x very large} ~ \ln x \simeq \frac{1}{1} + \frac{1}{2} +\frac{1}{3} +...+\frac{1}{x} [/tex]

    a result which can be recast rigorously in terms of limits.
     
  19. lanedance

    lanedance 3,307
    Homework Helper

    Hey sbcdave -welcome to PF!

    you're generally best off posting question as new threads, gets less confusing that way, particularly for old threads.

    The reminann integral (there are other more complex defintions) is interpreted as the area under of a well behaved function

    With this is mind and considering it as a definite integral (over a given integral where the function is well behaved
    [tex] \int_a^bdx \frac{1}{x} = ln(x)|_a^bdx =ln(b)-ln(a) [/tex]

    This is equivalent to the area between the function and the horizontal axis. Notice ln(x) is a a "monotonically" increasing function (ln(b)>ln(a) for all 0<a<b), so it always gives a positive area.

    I don't totally undertstand your question..

    But, at x = 0, f(x) = 1/x is not well defined, so to calculate the area you must use limits.

    Similarly to deal with taking the integral to infinity, you need to use a limiting process.

    Things get subtle when you consder lmiting process, so its important to be rigrorous. First lest split the integral into 2 portions, which we can do when the function is well behaved:
    [tex] \int_a^bdx f(x) = \int_a^cdx f(x) + \int_c^bdx f(x) [/tex]

    In this case lets choose c=1 and the integral becomes
    [tex] \lim_{a \to 0^+} \lim_{b \to +\infty} \int_a^b dx \frac{1}{x} = \lim_{a \to 0^+} \int_a^1 dx \frac{1}{x} + \lim_{b \to +\infty}\int_1^bdx \frac{1}{x} = \lim_{a \to 0^+}(ln(1)-ln(a)) + \lim_{b \to +\infty}(ln(b)-ln(1)) = \lim_{a \to 0^+}{-ln(a)} + \lim_{b \to +\infty}ln(b) [/tex]

    As both these diverge (become infinite) we actually find there is inifinite area below 1/x in both the intervals (0,1) and (1,inf). In fact in some repsects there is saimialr amount of infinte area in each.

    Though in both cases the curve compresses against the axis, it doesn't do so quickly enough to result in a finite area - things can get wierd with lmiting processes

    Hopefully this helped answer your question and didn't confuse things more!

    can be a nice way to help, learn and keep up with some math skills

    Mod note: fixed LaTeX
     
    Last edited by a moderator: Dec 21, 2011
  20. Thanks
    Roger
    But the natural log of anything less than 1 is negative, which seems to say that the area under the curve from 0 to anything less than x=1 on the graph of f(x)=1/x should be a negative area. The graph of f(x)=1/x from x=0 to 1 is above the x-axis and as you say in the next quote is infinite.

    Example of what I'm struggling with: [tex] \int_{0.1}^1dx \frac{1}{x} [/tex]

    Should be a definable positive integer. Approximately equal to 9(0.1) increments with heights 1/0.9 + 1/0.8 + ... 1/0.1 (left hand method) or approximately equal to 2.829. However, if you do ln(1)-ln(0.1) you get approximately 2.3 (I expected a negative number and need to think about this a little more, there's clearly a flaw in my logic...haha) Sorry. Posting anyways so that you might see where I was coming from.
     
  21. I was looking at a graph of ln(x) and 1/x together on my ti-83 and assumed that because the ln(x) went negative with x<1 that that meant the area was negative, now I see that because the -y value grows as x gets smaller that your subtracting a larger negative from a negative and end up with a positive.

    Now I'm a little confused why the left hand method was larger than the ln(x)-ln(x) method instead of smaller though if you have any ideas on that?

    Thanks again for humoring me once already I'll understand if the thread dies here...lol
     
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