Question

Help please! need the integral from -infinity to +infinity of 1/(x^2+1)

Answer 1

i know i would use the tanx=u sub and

Answer 2

wouldnt it be 0

Answer 3

\(\int_{-\infty}^\infty\frac{1}{x^2+1}dx=\tan^{-1}(x)|_{-\infty}^{\infty}=\frac{\pi}{2}-(-\frac{\pi}{2})=\pi\)

Answer 4

It can't be 0 because the integrand is always positive.

Answer 5

\(\lim_{x\rightarrow \infty}\tan^{-1}(x)=\frac{\pi}{2}\\\lim_{x\rightarrow -\infty}\tan^{-1}(x)=-\frac{\pi}{2}\)

Answer 6

\[\int\limits_{-\infty}^{0} \tan^-1x +\int\limits_{0}^{\infty} \tan^-1\]

Answer 7

is that ok?

Answer 8

and then do the lim as b aproaches infinity and neg and etc

Answer 9

no arctan is what you get after you take the integral

Answer 10

ok

Answer 11

but can i break the intrgrand like that?

Answer 12

\(\lim_{t\rightarrow-\infty}(-\int_{0}^t\frac{1}{x^2+1}dx)+\lim_{t_0\rightarrow\infty}\int_0^{t_0}\frac{1}{x^2+1}dx\)

Answer 13

u can prob use t instead of t_0 in the second term, but ....just to be safe

Answer 14

ok that perf ok thanks!

Answer 15

thats how i set it up