Originally Posted by
earboth 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
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