Changes

The reason we chose this contour is that it necessarily avoids the inevitable singularity of the logarithm at 0, and it includes inside of it, the pole of the function <math>f(z)=log(z)/(1+z^2)</math>. By Cauchy's residue theorem, we have <math>I:=\int_{\Gamma(r,R)}f(z)\,dz</math>= <math>2\pi i</math> times the residue of the [[pole]] at <math>z=i</math>. This is easily calculated, and it is equal to <math>\pi/4</math>. Thus, the integral evaluates to <math>\pi^2 i/2</math>.

The reason we chose this contour is that it necessarily avoids the inevitable singularity of the logarithm at 0, and it includes inside of it, the pole of the function <math>f(z)=log(z)/(1+z^2)</math>. By Cauchy's residue theorem, we have <math>I:=\int_{\Gamma(r,R)}f(z)\,dz</math>= <math>2\pi i</math> times the residue of the [[pole]] at <math>z=i</math>. This is easily calculated, and it is equal to <math>\pi/4</math>. Thus, the integral evaluates to <math>\pi^2 i/2</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>.

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]]

[[Category:integration]]

[[Category:integration]]