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[tex]\int\frac{dx}{1+x^{2}} [/tex] (1) where the integration is from -infinity to +infinity.

To do this, we must, apparently consider:

[tex]\oint\frac{dz}{1+z^2}[/tex] (2). The closed loop is a countour which is a semi circle of radius R about the origin containing ONLY the +i singularity (there are two singularities, +i and -i). .

Ofcourse we may write [tex]\frac{1}{1+z^2}=-\frac{1}{2i}\left(\frac{1}{z+i}-\frac{1}{z-i}\right)[/tex]

Thus the residue at z=i is 1/2i. Therefore the integral (2) is 2pi*i*res(z=i)=pi

This implies (I can't understand why) the Integral (1) EQUALS Integral (2)=pi.

Questions:

1)why do we consider only a loop containing one of the singularities (+i)?

2)why is integral (1), with limits going from -infinity to plus infinity EQUAL to the same integral, with weird limits (-R to R along a semi circle, then along real axis from -R to R)?

Thanks in advance:)