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Why is the integral of 1/x from -1 to 1 divergent?

  1. Jan 6, 2016 #1
    Hey guys, could someone help explain to me why the integral of 1/x from -1 to 1 is considered divergent? It would seem as if the area underneath the function cancels out with each other to give you the result of zero, but apparently this is not the case.

    Thanks!
     
  2. jcsd
  3. Jan 6, 2016 #2

    andrewkirk

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    The integral does not exist because the interval [-1,1] does not lie in the domain of the function. That's because 0 is excluded from the domain. So the expression ##\int_{-1}^1\frac{dx}{x}## is meaningless.
    The closest you can come is to calculate
    $$g(x)\equiv\lim_{\delta\to 0}\left[\int_{-1}^{-\delta}\frac{dx}{x}+\int_{\delta}^{1}\frac{dx}{x}\right]$$
    which will be zero. But that does not match the definition of an integral on [-1,1] because, as stated above, the integral is not defined.
     
  4. Jan 6, 2016 #3

    pwsnafu

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    It is the difference between an improper integral and a principal value integral.

    In the former, we need to evaluate
    ##\lim_{x\to0} \int_{-1}^{x} \frac{1}{t} dt + \lim_{y\to0} \int_{y}^1 \frac{1}{t}dt##.
    Notice that x and y are independent of each other. If you take the limit of x faster* than y, you will evaluate ##-\infty##. If you let y converge faster than x, you will evaluate ##\infty##. These don't match hence the integral is undefined.

    With PV integrals we need eavluate
    ##\lim_{\epsilon \to 0} \left(\int_{-1}^{-\epsilon} \frac{1}{t}dt + \int_{\epsilon}^{1}\frac{1}{t}dt \right)##
    Notice how they converge from both sides at the same rate. Hence the two integrals always cancel leaving zero.

    Unless you explicitly state principal value, the default is to assume improper as it is more useful in applications.

    * Aside: you should try to think of limits as a process that converges to anything, but in the this example it works.
     
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