# Diverge or Converge

• Nov 4th 2006, 06:49 AM
FLTR
Diverge or Converge
Determine whether $\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx$ converges or diverges. If it converges, evaluate the integral.
• Nov 4th 2006, 07:07 AM
earboth
Quote:

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

Hello,FLTR,

this is a standard integral which you probably know:

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

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

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

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

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 PM
topsquark
Quote:

Originally Posted by earboth
Hello,FLTR,

this is a standard integral which you probably know:

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

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

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

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

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!

I don't see anything wrong with the Math and my calculator, at least, agrees with you.

-Dan
• Nov 4th 2006, 01:58 PM
ThePerfectHacker
Since,
$f(x)=\frac{1}{1+x^2}$
Is continous on the entire number line, we well-define

$\int_{-\infty}^{\infty}$
At any point,
Choosing $x=0$

$\int_{-\infty}^0+\int_0^{\infty}$
Both the improper integrals converge to what Earboth said.