Changes

The idea is to now take <math>r\rightarrow 0^+</math> and <math>R\rightarrow\infty</math>. Some easy calculations (left to the reader) show that the integrals along the semicircular arcs vanish, leaving us with only the integrals along the [[real axes]]. For <math>(0,\infty)</math> we get back the integral we are trying to calculate <math>I</math>, and on <math>(-\infty,0)</math> we get (since arg here is equal to <math>\pi i</math>): <math>I+\pi i\int_{-\infty}^0 dz/(1+z^2)=I+\pi^2 i/2</math>. Thus, plugging everything back in, <math>2I+\pi^2 i/2=\pi^2 i/2</math>, so <math>I=0</math>.

The idea is to now take <math>r\rightarrow 0^+</math> and <math>R\rightarrow\infty</math>. Some easy calculations (left to the reader) show that the integrals along the semicircular arcs vanish, leaving us with only the integrals along the [[real axes]]. For <math>(0,\infty)</math> we get back the integral we are trying to calculate <math>I</math>, and on <math>(-\infty,0)</math> we get (since arg here is equal to <math>\pi i</math>): <math>I+\pi i\int_{-\infty}^0 dz/(1+z^2)=I+\pi^2 i/2</math>. Thus, plugging everything back in, <math>2I+\pi^2 i/2=\pi^2 i/2</math>, so <math>I=0</math>.

[[Category:Calculus]]

[[Category:Calculus]]