miércoles, 24 de julio de 2019

integrals circulars de límit zero

lim[s-->0]int[z=se^{xi}+a][ f(z)/(z+(-a)) ]d[z]·(1/2pi) = f(a)
z=se^{xi}+a


lim[s-->0]int[( ln(z) )^{n}=se^{xi}+a][ f(z)/( ( ln(z) )^{n}+(-a) ) ]d[z]·(1/2pi) = ...
...= f( e^{a^{(1/n)}} )·e^{a^{(1/n)}}·(1/n)·a^{((1/n)+(-1))}
z=e^{( se^{xi}+a )^{(1/n)}}


lim[s-->0]int[( ln(z)+c )^{n}=se^{xi}+a][ f(z)/( ( ln(z)+c )^{n}+(-a) ) ]d[z]·(1/2pi) = ...
...= f( e^{a^{(1/n)}+(-c)} )·e^{a^{(1/n)+(-c)}}·(1/n)·a^{((1/n)+(-1))}
z=e^{( se^{xi}+a )^{(1/n)}+(-c)}


lim[s-->0]int[( e^{x} )^{n}=se^{xi}+a][ f(z)/( ( e^{x} )^{n}+(-a) ) ]d[z]·(1/2pi) = ...
...= f( ln( a^{(1/n)} ) )·(1/n)·(1/a)
z=ln( ( se^{xi}+a )^{(1/n)} )


lim[s-->0]int[( e^{x}+c )^{n}=se^{xi}+a][ f(z)/( ( e^{x}+c )^{n}+(-a) ) ]d[z]·(1/2pi) = ...
...= f( ln( a^{(1/n)}+(-c) ) )·( 1/(a^{(1/n)}+(-c)) )·(1/n)·a^{(1/n)+(-1))}
z=ln( ( se^{xi}+a )^{(1/n)}+(-c) )


lim[s-->0]int[( x+c )^{n}=se^{xi}+a][ f(z)/( ( x+c )^{n}+(-a) ) ]d[z]·(1/2pi) = ...
...= f( a^{(1/n)}+(-c) )·(1/n)·a^{(1/n)+(-1))}
z=( se^{xi}+a )^{(1/n)}+(-c)


lim[s-->0]int[( x^{p}+c )^{n}=se^{xi}+a][ f(z)/( ( x^{p}+c )^{n}+(-a) ) ]d[z]·(1/2pi) = ...
...= f( ( a^{(1/n)}+(-c) )^{(1/p)} )·(1/p)·(a^{(1/n)+(-c)})^{((1/p)+(-1))}·(1/n)·a^{(1/n)+(-1))}
z=( ( se^{xi}+a )^{(1/n)}+(-c) )^{(1/p)}

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