Determine whether $\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx$ converges or diverges. If it converges, evaluate the integral.

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- Nov 4th 2006, 06:49 AMFLTRDiverge or Converge
Determine whether $\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx$ converges or diverges. If it converges, evaluate the integral.

- Nov 4th 2006, 07:07 AMearboth
Hello,FLTR,

this is a standard integral which you probably know:

$\displaystyle \int \frac{1}{1+x^2}dx=\arctan(x)+C$

As you may know the arctan-function has the limit pi/2 if x approaches infinity:

$\displaystyle \lim_{x\rightarrow \infty} \arctan(x)=\frac{\pi}{2}$ and

$\displaystyle \lim_{x\rightarrow -\infty} \arctan(x)=-\frac{\pi}{2}$

So you get: $\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}-\left( -\frac{\pi}{2} \right) = \pi$

EB

Unfortunately the preview doesn't work properly. So I send you this post without checking. Please be careful, there might be some ugly mistakes! - Nov 4th 2006, 12:43 PMtopsquark
- Nov 4th 2006, 01:58 PMThePerfectHacker
Since,

$\displaystyle f(x)=\frac{1}{1+x^2}$

Is continous on the entire number line, we well-define

$\displaystyle \int_{-\infty}^{\infty}$

At any point,

Choosing $\displaystyle x=0$

$\displaystyle \int_{-\infty}^0+\int_0^{\infty}$

Both the improper integrals converge to what Earboth said.