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 2 years ago
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(x1)^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
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(x1)^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

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akshay4
 2 years ago
Best ResponseYou've already chosen the best response.0convergent bcoz integral will come out to b 1 (finite)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2the lower bound is also not defined, so it has two undefined bounds, hence it is twice improper

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2also\[\int\frac{dx}{(x1)^2}\ne \ln(1x)^2\]

xillahah
 2 years ago
Best ResponseYou've already chosen the best response.1are there certain steps that i must take just because it is twice improper or do i just go about solving the integral normally?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2you integrate normally, but you will have to take the limits for both bounds\[\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^m\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2\[\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^mf(x)dx\]

xillahah
 2 years ago
Best ResponseYou've already chosen the best response.1i 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 = x1 would give me (x1)^1 giving me 1/x1 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2your 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2hence you rewrite this\[\lim_{n\to\infty}\lim_{m\to1}\left.\frac1{x1}\right_n^m\]now evaluate and take the limits

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2I mean\[\lim_{n\to\infty}\lim_{m\to1}\left.\frac1{x1}\right_m^n\]

xillahah
 2 years ago
Best ResponseYou've already chosen the best response.1or separating these limits would give me lim n> infinity 1/x1 = 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2correct, it diverges
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