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xillahah Group Title

integral from one to infinity (dx)/(x - 1)^2 ..is it convergent or divergent? my prof tells me it is a twice improper integral..so i don't know what he means by that. when i first attempted the problem i just took the ln of the integral giving me ln(x-1)^2 with b = infinity and lower bound being 1. after evaluating the integral i came up with an answer of infinity. making the equation divergent

  • 2 years ago
  • 2 years ago

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  1. akshay4 Group Title
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    convergent bcoz integral will come out to b 1 (finite)

    • 2 years ago
  2. xillahah Group Title
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    how though?

    • 2 years ago
  3. TuringTest Group Title
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    the lower bound is also not defined, so it has two undefined bounds, hence it is twice improper

    • 2 years ago
  4. TuringTest Group Title
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    also\[\int\frac{dx}{(x-1)^2}\ne \ln(1-x)^2\]

    • 2 years ago
  5. xillahah Group Title
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    are there certain steps that i must take just because it is twice improper or do i just go about solving the integral normally?

    • 2 years ago
  6. TuringTest Group Title
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    you integrate normally, but you will have to take the limits for both bounds\[\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^m\]

    • 2 years ago
  7. TuringTest Group Title
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    \[\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^mf(x)dx\]

    • 2 years ago
  8. xillahah Group Title
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    i took u = x - 1 and my du = dx. so..\[\int\limits_{?}^{?}du/u^2\] then bringing up the u^2 would make it u^-2 du giving me \[-u ^{-1}\]..plugging in u = x-1 would give me -(x-1)^-1 giving me -1/x-1 as an answer to the integral right? if so, i don't understand what you mean by taking limits for both bounds. what would my bounds be for infinity and 1? |dw:1352148781734:dw|

    • 2 years ago
  9. TuringTest Group Title
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    correct

    • 2 years ago
  10. TuringTest Group Title
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    your bounds are infinity and one, but technically you can't just plug those in, since infinity is not a number, and the integral is not defined at x=1

    • 2 years ago
  11. TuringTest Group Title
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    hence you rewrite this\[\lim_{n\to\infty}\lim_{m\to1}\left.-\frac1{x-1}\right|_n^m\]now evaluate and take the limits

    • 2 years ago
  12. TuringTest Group Title
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    I mean\[\lim_{n\to\infty}\lim_{m\to1}\left.-\frac1{x-1}\right|_m^n\]

    • 2 years ago
  13. xillahah Group Title
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    or separating these limits would give me lim n-> infinity -1/x-1 = 0 and lim m-> 1 that would give me 1/0 resulting in the limit not existing then the whole entire thing would be divergent

    • 2 years ago
  14. TuringTest Group Title
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    correct, it diverges

    • 2 years ago
  15. xillahah Group Title
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    thank you so much!

    • 2 years ago
  16. TuringTest Group Title
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    welcome!

    • 2 years ago
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