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B Integrating to infinity issue

  1. May 29, 2017 #1
    Say we have two functions; ##f\left(x\right)=\frac{1}{e^x-1}## and ##g\left(x\right)=\ln \left(\frac{1}{x}+1\right)##. Let us find the limit of both functions as x approaches infinity;
    ##\lim_{x \rightarrow \infty} {f(x)} = \frac{1}{e^\infty-1} = \frac{1}{\infty} = 0## Therefore as ##x \rightarrow \infty##, ##f(x) \rightarrow 0##.
    ##\lim_{x \rightarrow \infty} {g(x)} = \ln \left(\frac{1}{\infty}+1\right) = \ln \left(0+1\right) = 0## Therefore as ##x \rightarrow \infty##, ##g(x) \rightarrow 0##.

    Now, when we integrate ##f(x)## over the domain ##[1, \infty)##, it works perfectly fine and has a result of
    0.45868. This makes sense because the area enclosed under the graph is constantly getting smaller due to ##f(x) \rightarrow 0##.

    However, when I try to integrate ##g(x)## over the same domain my calculator wont give me an answer. ##g(x)## approaches 0 just as ##f(x)## does, meaning that the enclosed area should increasingly get smaller also until it reaches the point where the area incriminates being added become negligible. So why can't I get a numerical answer while integrating ##g(x)##?
     
  2. jcsd
  3. May 30, 2017 #2

    Orodruin

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    It is not sufficient that the function goes to zero. It must go to zero sufficiently fast.
     
  4. May 30, 2017 #3
    Could you please elaborate on what you mean? How could the function not being 'sufficiently fast' change the fact that it approaches zero as x approaches infinity? Integrating gives the area under the graph, and we know that as x approaches infinity the function approaches 0. This means there has to be a point in which the 'height' of the function above the x-axis becomes negligible (that is equal to zero) and therefore the area under the the graph must stop having incriminates added onto it.
     
  5. May 30, 2017 #4

    PeroK

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    A function tending to 0 is not the same as a function becoming 0. Your functions never actually reach 0, hence there are always"increments".

    Look up the series ##\Sigma 1/n## for a clue as to what is going on here.
     
  6. May 30, 2017 #5

    Orodruin

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    No it doesn't. This argument is not rigorous. Even if the function value decreases towards zero, it does not mean that the area will. Consider the piecewise constant function that takes the value ##1/2^n## between ##x = 2^n## and ##x = 2^{n+1}##. Clearly, this function converges to zero as ##x \to \infty##. However, integrating the function from ##2^n## to ##2^{n+1}## gives ##2^n/2^n = 1##. Therefore, the integral from 1 to ##2^n## is equal to ##n## and therefore the integral diverges when taking the upper limit to infinity. When you compute an area, it is not only a matter of "height", you also need to consider the width of the region and your region is infinite.
     
  7. May 30, 2017 #6
    I'm sorry for being so dense everyone, I haven't even heard of things like piecewise constants. I guess I'm just beyond my current level of understanding here.

    There is one thing I stumbled across though - the derivative function of ##g\left(x\right)=\ln \left(\frac{1}{x}+1\right)## is ##g'(x)=\frac{-1}{x(x+1)}##. The squared factor on the denominator implies that; as x approaches infinity, the gradient function approaches 0 at a faster rate than function ##g(x)## approaches 0. Could you then extrapolate from this and state that ##g(x)## approaches a non-zero constant value before approaching 0, and thus the area under the graph will be infinite?
     
  8. May 30, 2017 #7

    Orodruin

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    No.

    Think of it this way, if the integrant approached a constant non-zero value ##c_0## and the integral upper boundary was set to ##x = N##, then the integral would grow as ##\sim c_0 N## as ##N \to \infty##. However, if the integrand goes to zero, the integral will increase slower. However, there are many possible functions that grow slower than ##N## as ##N \to \infty## but that still tend to infinity, ##\log(N)## being one such example.
     
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