Contour Integrals involving Log(-x), where does negative sign go?
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beginalign -i pi^2 &= left( int_R + int_M + int_N + int_r right) f(z) , dz \
&= left( int_M + int_N right) f(z), dz && int_R, int_r mbox vanish \
&=-int_infty^0 left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2, dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2right)^2 , dx \
&= int_0^infty left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2 , dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2 right )^2 , dx \
&= int_0^infty left (fraclog x + ipi1+x^2 right )^2 , dx - int_0^infty left (fraclog x - ipi1+x^2 right )^2 , dx && varepsilon to 0 \
&= int_0^infty frac(log x + ipi)^2 - (log x - ipi)^2left(1+x^2right)^2 , dx \
&= int_0^infty frac4 pi i log xleft(1+x^2right)^2 , dx \
&= 4 pi i int_0^infty fraclog xleft(1+x^2right)^2 , dx endalign
When epsilon is taken to zero, what is the argument that lets us ignore that there's a negative sign in front of x? Where does it go? We go to polar representation implicitly?
logarithms contour-integration cauchy-integral-formula
asked Sep 10 at 19:21
walczyk
5119
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beginalign -i pi^2 &= left( int_R + int_M + int_N + int_r right) f(z) , dz \
&= left( int_M + int_N right) f(z), dz && int_R, int_r mbox vanish \
&=-int_infty^0 left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2, dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2right)^2 , dx \
&= int_0^infty left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2 , dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2 right )^2 , dx \
&= int_0^infty left (fraclog x + ipi1+x^2 right )^2 , dx - int_0^infty left (fraclog x - ipi1+x^2 right )^2 , dx && varepsilon to 0 \
&= int_0^infty frac(log x + ipi)^2 - (log x - ipi)^2left(1+x^2right)^2 , dx \
&= int_0^infty frac4 pi i log xleft(1+x^2right)^2 , dx \
&= 4 pi i int_0^infty fraclog xleft(1+x^2right)^2 , dx endalign
When epsilon is taken to zero, what is the argument that lets us ignore that there's a negative sign in front of x? Where does it go? We go to polar representation implicitly?
logarithms contour-integration cauchy-integral-formula
asked Sep 10 at 19:21
walczyk
5119
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
beginalign -i pi^2 &= left( int_R + int_M + int_N + int_r right) f(z) , dz \
&= left( int_M + int_N right) f(z), dz && int_R, int_r mbox vanish \
&=-int_infty^0 left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2, dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2right)^2 , dx \
&= int_0^infty left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2 , dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2 right )^2 , dx \
&= int_0^infty left (fraclog x + ipi1+x^2 right )^2 , dx - int_0^infty left (fraclog x - ipi1+x^2 right )^2 , dx && varepsilon to 0 \
&= int_0^infty frac(log x + ipi)^2 - (log x - ipi)^2left(1+x^2right)^2 , dx \
&= int_0^infty frac4 pi i log xleft(1+x^2right)^2 , dx \
&= 4 pi i int_0^infty fraclog xleft(1+x^2right)^2 , dx endalign
When epsilon is taken to zero, what is the argument that lets us ignore that there's a negative sign in front of x? Where does it go? We go to polar representation implicitly?
logarithms contour-integration cauchy-integral-formula
asked Sep 10 at 19:21
walczyk
5119
beginalign -i pi^2 &= left( int_R + int_M + int_N + int_r right) f(z) , dz \
&= left( int_M + int_N right) f(z), dz && int_R, int_r mbox vanish \
&=-int_infty^0 left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2, dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2right)^2 , dx \
&= int_0^infty left (fraclog(-x + ivarepsilon)1+(-x + ivarepsilon)^2 right )^2 , dx - int_0^infty left (fraclog(-x - ivarepsilon)1+(-x - ivarepsilon)^2 right )^2 , dx \
&= int_0^infty left (fraclog x + ipi1+x^2 right )^2 , dx - int_0^infty left (fraclog x - ipi1+x^2 right )^2 , dx && varepsilon to 0 \
&= int_0^infty frac(log x + ipi)^2 - (log x - ipi)^2left(1+x^2right)^2 , dx \
&= int_0^infty frac4 pi i log xleft(1+x^2right)^2 , dx \
&= 4 pi i int_0^infty fraclog xleft(1+x^2right)^2 , dx endalign
When epsilon is taken to zero, what is the argument that lets us ignore that there's a negative sign in front of x? Where does it go? We go to polar representation implicitly?
logarithms contour-integration cauchy-integral-formula
logarithms contour-integration cauchy-integral-formula
asked Sep 10 at 19:21
walczyk
5119
asked Sep 10 at 19:21
walczyk
5119
asked Sep 10 at 19:21
walczyk
5119
asked Sep 10 at 19:21
walczyk
5119
asked Sep 10 at 19:21
walczyk
5119
5119
add a comment |Â
add a comment |Â
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