xillahah 3 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(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

1. akshay4

convergent bcoz integral will come out to b 1 (finite)

2. xillahah

how though?

3. TuringTest

the lower bound is also not defined, so it has two undefined bounds, hence it is twice improper

4. TuringTest

also$\int\frac{dx}{(x-1)^2}\ne \ln(1-x)^2$

5. xillahah

are there certain steps that i must take just because it is twice improper or do i just go about solving the integral normally?

6. TuringTest

you integrate normally, but you will have to take the limits for both bounds$\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^m$

7. TuringTest

$\int_a^bf(x)dx=\lim_{n\to a}\lim_{m\to b}\int_n^mf(x)dx$

8. xillahah

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|

9. TuringTest

correct

10. TuringTest

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

11. TuringTest

hence you rewrite this$\lim_{n\to\infty}\lim_{m\to1}\left.-\frac1{x-1}\right|_n^m$now evaluate and take the limits

12. TuringTest

I mean$\lim_{n\to\infty}\lim_{m\to1}\left.-\frac1{x-1}\right|_m^n$

13. xillahah

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

14. TuringTest

correct, it diverges

15. xillahah

thank you so much!

16. TuringTest

welcome!