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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
 one year ago
 one year 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
 one year ago
 one year ago

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akshay4Best ResponseYou've already chosen the best response.0
convergent bcoz integral will come out to b 1 (finite)
 one year ago

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

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

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

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

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

xillahahBest ResponseYou've already chosen the best response.1
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 = 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

xillahahBest ResponseYou've already chosen the best response.1
or 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
correct, it diverges
 one year ago
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