Question

determine if the doubly infinite improper integral converges and, if so, evaluate it determine if the doubly infinite improper integral converges and, if so, evaluate it @Mathematics

Answer 1

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

Answer 2

wait this not math

Answer 3

wrong grup

Answer 4

yea it is...

Answer 5

sorry

Answer 6

so we need to break up the integral
i'm going to choose the constant to be 0
so we have
\[\int\limits_{-\infty}^{0}\frac{1}{1+x^2} dx +\int\limits_{0}^{\infty}\frac{1}{1+x^2} dx\]

Answer 7

well now i feel dumb

Answer 8

The infamous arctan(x) function. We meet again.

Answer 9

\[\lim_{a \rightarrow - \infty}[\tan^{-1}(x)]_{a}^0 +\lim_{b \rightarrow \infty}[\tan^{-1}(x)]_0^{b}\]

Answer 10

\[\lim_{a \rightarrow -\infty}[\tan^{-1}(0)-\tan^{-1}(a)]+\lim_{b \rightarrow \infty}[\tan^{-1}(b)-\tan^{-1}(0)]\]

Answer 11

\[0-\lim_{a \rightarrow -\infty} \tan^{-1}(a)+\lim_{b \rightarrow \infty}\tan^{-1}(b)-0\]
\[0-\frac{-\pi}{2}+\frac{\pi}{2}-0\]
so both integrals are convergent so that means the one we started out with his convergent
and actually converges to pi

Answer 12

thank you