Diverge or Converge

**FLTR** · November 4th 2006, 06:49 AM

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

**earboth** · November 4th 2006, 07:07 AM

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!

**topsquark** · November 4th 2006, 12:43 PM

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

**ThePerfectHacker** · November 4th 2006, 01:58 PM

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.

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