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Math Help - Definite Integral

  1. #1
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    Definite Integral

    I am taking the integral of a function that contains a discontinuity. So in order to do that, I first take the integral from 0 to 1, and from 1 to 2 and add the two integrals. Here's my problem: the function, once integrated, is ln|(x-1)|+x. So I have [ ln|(1-1)|+1 ] - [ ln|(0-1)|+0 .

    ln(0) is - infinity? How do I evaluate the definite integral?
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    Quote Originally Posted by onlinemathstudent View Post
    I am taking the integral of a function that contains a discontinuity. So in order to do that, I first take the integral from 0 to 1, and from 1 to 2 and add the two integrals. Here's my problem: the function, once integrated, is ln|(x-1)|+x. So I have [ ln|(1-1)|+1 ] - [ ln|(0-1)|+0 .

    ln(0) is - infinity? How do I evaluate the definite integral?
    What is the integral you are trying to integrate?
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  3. #3
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    f(x) = (x^2+x) / (x^2 -1)
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    Quote Originally Posted by onlinemathstudent View Post
    I am taking the integral of a function that contains a discontinuity. So in order to do that, I first take the integral from 0 to 1, and from 1 to 2 and add the two integrals. Here's my problem: the function, once integrated, is ln|(x-1)|+x. So I have [ ln|(1-1)|+1 ] - [ ln|(0-1)|+0 .

    ln(0) is - infinity? How do I evaluate the definite integral?
    is this the original integral?

    \displaystyle \int_0^1 \frac{1}{x-1} + 1 \, dx

    if so, it is an improper integral that diverges ... in other words, the following limit does not exist.

    \displaystyle \lim_{b \to 1^-} \int_0^b \frac{1}{x-1} + 1 \, dx
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    Here's the whole problem:

    Definite Integral-screenshot-2.png
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  6. #6
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    Ok. After a little research into the matter, I see that there's no way to do this. It is an INFINITE discontinuity and therefore cannot be integrated. Had it been a JUMP discontinuity, then I cou;d have done the two integrals.
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