Question

Improper Integrals: show that the integral from 1 to infinity of 1/(1+x^2) dx is divergent.

Answer 1

integrate to get...tanx...add limx->inf as this does not exist it diverges

Answer 2

I did this knowing that the integral of 1/(1+x^2) is arctan, but that is where I am hitting trouble. Plug in the limits and use FTC, I get pi/2 and 45 (degrees).

Answer 3

\[\int\limits_{1}^{\infty}dx/(x^2+1)=\tan^{-1} |[1,\infty]\]

since \[\tan^{-1} (\infty)=\pi/2\]
and\[\tan^{-1} (1)=45^* or \pi/4\]

pi/2-pi/4=pi/4 on a good day to me

Answer 4

I got this, so how does it diverge? To my knowledge, if you get a finite answer, then the integral converges....The problem asks to show that it diverges.

Answer 5

Maybe I am just over thinking what the question is asking. IDK thanks for your help though.

Answer 6

let infinity be t the do the lim as t go to infinity of the same integral and change the bounds to [1,t] and you get tan^-1(t)-tan^-1(1)= infinity because tan^-1(infinty) never hits pi/2 y is going to infinity and theorfore divergent. try this with 1/x integral from 1 to infinty and you will get infinity or divergent as well. hope that helped if i had my book i would look it over but i dont

Answer 7

OK, I understand now. Was confused with arctan infinity = pi/2, but you clarified when you said it never reaches. Thanks again.

Answer 8

It isn't divergent, it's\[\lim_{t \to \infty} \arctan(t) - \arctan(1) = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}.\]

Answer 9

so it is...it isn't...now I am lost.

Answer 10

It isn't. bnut056's first reply is correct, and it shows it's\[\pi/4.\]

http://www.wolframalpha.com/input/?i=integrate+1%2F%281%2Bx^2%29+from+x+%3D+1+to+inf

Answer 11

smooootthhhhh!!! Much appreciated.

Answer 12

I KNEW IT, ITS COVERGENT i need to keep my book near me when doing these problems