integral with residues , tangent and sin

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Hi guys do you know how i can solve this integral ?
enter image description hereI can't use Laurent series.
The problem is that i don't know how i can solve the tan^2 part. in the denominator we have $z=fracpi2$ but it's a zero als for the numerator ( but order 2) , so z is a zero of order 1 for the denominator and order 2 for the numerator... but what about the tan part ? how i have to do that ?






integration complex-analysis residue-calculus







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  • $pi / 2$ is a pole of order 3
    – Davide Morgante
    Sep 3 at 13:28










  • why ? (1-sinz) get the order 2 and the 2z-pi is order 1 so it should not be a pole
    – xmaionx
    Sep 3 at 13:30










  • $operatornameord_z_0(fg) = operatornameord_z_0(f)+operatornameord_z_0(g)$
    – Davide Morgante
    Sep 3 at 13:32










  • ah but how i can solve the tan part ?
    – xmaionx
    Sep 3 at 13:37











  • You can evaluate residues of higher order by derivation: e.g. $z=pi/2$ is a order 3 pole. The Laurent expansion of $$f(z) = fraca_-3(z-fracpi2)^3+ fraca_-2(z-fracpi2)^2+ fraca_-1(z-fracpi2)+a_0+cdots$$ so $$fracd^2dz^2(f(z)(z-fracpi2)^3)= fracd^2dz^2left(a_-3+(z-fracpi2)a_-2+(z-fracpi2)^2a_-1+cdotsright)$$ taking the limit as $zrightarrow fracpi2$, the second derivative gives you $2a_-1$ which is two times the residue in $fracpi2$
    – Davide Morgante
    Sep 3 at 14:02















up vote
-4
down vote

favorite
1












Hi guys do you know how i can solve this integral ?
enter image description hereI can't use Laurent series.
The problem is that i don't know how i can solve the tan^2 part. in the denominator we have $z=fracpi2$ but it's a zero als for the numerator ( but order 2) , so z is a zero of order 1 for the denominator and order 2 for the numerator... but what about the tan part ? how i have to do that ?






integration complex-analysis residue-calculus







share|cite|improve this question























  • $pi / 2$ is a pole of order 3
    – Davide Morgante
    Sep 3 at 13:28










  • why ? (1-sinz) get the order 2 and the 2z-pi is order 1 so it should not be a pole
    – xmaionx
    Sep 3 at 13:30










  • $operatornameord_z_0(fg) = operatornameord_z_0(f)+operatornameord_z_0(g)$
    – Davide Morgante
    Sep 3 at 13:32










  • ah but how i can solve the tan part ?
    – xmaionx
    Sep 3 at 13:37











  • You can evaluate residues of higher order by derivation: e.g. $z=pi/2$ is a order 3 pole. The Laurent expansion of $$f(z) = fraca_-3(z-fracpi2)^3+ fraca_-2(z-fracpi2)^2+ fraca_-1(z-fracpi2)+a_0+cdots$$ so $$fracd^2dz^2(f(z)(z-fracpi2)^3)= fracd^2dz^2left(a_-3+(z-fracpi2)a_-2+(z-fracpi2)^2a_-1+cdotsright)$$ taking the limit as $zrightarrow fracpi2$, the second derivative gives you $2a_-1$ which is two times the residue in $fracpi2$
    – Davide Morgante
    Sep 3 at 14:02













up vote
-4
down vote

favorite
1









up vote
-4
down vote

favorite
1






1





Hi guys do you know how i can solve this integral ?
enter image description hereI can't use Laurent series.
The problem is that i don't know how i can solve the tan^2 part. in the denominator we have $z=fracpi2$ but it's a zero als for the numerator ( but order 2) , so z is a zero of order 1 for the denominator and order 2 for the numerator... but what about the tan part ? how i have to do that ?






integration complex-analysis residue-calculus







share|cite|improve this question















Hi guys do you know how i can solve this integral ?
enter image description hereI can't use Laurent series.
The problem is that i don't know how i can solve the tan^2 part. in the denominator we have $z=fracpi2$ but it's a zero als for the numerator ( but order 2) , so z is a zero of order 1 for the denominator and order 2 for the numerator... but what about the tan part ? how i have to do that ?






integration complex-analysis residue-calculus




integration complex-analysis residue-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 3 at 15:10

























asked Sep 3 at 13:22









xmaionx

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  • $pi / 2$ is a pole of order 3
    – Davide Morgante
    Sep 3 at 13:28










  • why ? (1-sinz) get the order 2 and the 2z-pi is order 1 so it should not be a pole
    – xmaionx
    Sep 3 at 13:30










  • $operatornameord_z_0(fg) = operatornameord_z_0(f)+operatornameord_z_0(g)$
    – Davide Morgante
    Sep 3 at 13:32










  • ah but how i can solve the tan part ?
    – xmaionx
    Sep 3 at 13:37











  • You can evaluate residues of higher order by derivation: e.g. $z=pi/2$ is a order 3 pole. The Laurent expansion of $$f(z) = fraca_-3(z-fracpi2)^3+ fraca_-2(z-fracpi2)^2+ fraca_-1(z-fracpi2)+a_0+cdots$$ so $$fracd^2dz^2(f(z)(z-fracpi2)^3)= fracd^2dz^2left(a_-3+(z-fracpi2)a_-2+(z-fracpi2)^2a_-1+cdotsright)$$ taking the limit as $zrightarrow fracpi2$, the second derivative gives you $2a_-1$ which is two times the residue in $fracpi2$
    – Davide Morgante
    Sep 3 at 14:02

















  • $pi / 2$ is a pole of order 3
    – Davide Morgante
    Sep 3 at 13:28










  • why ? (1-sinz) get the order 2 and the 2z-pi is order 1 so it should not be a pole
    – xmaionx
    Sep 3 at 13:30










  • $operatornameord_z_0(fg) = operatornameord_z_0(f)+operatornameord_z_0(g)$
    – Davide Morgante
    Sep 3 at 13:32










  • ah but how i can solve the tan part ?
    – xmaionx
    Sep 3 at 13:37











  • You can evaluate residues of higher order by derivation: e.g. $z=pi/2$ is a order 3 pole. The Laurent expansion of $$f(z) = fraca_-3(z-fracpi2)^3+ fraca_-2(z-fracpi2)^2+ fraca_-1(z-fracpi2)+a_0+cdots$$ so $$fracd^2dz^2(f(z)(z-fracpi2)^3)= fracd^2dz^2left(a_-3+(z-fracpi2)a_-2+(z-fracpi2)^2a_-1+cdotsright)$$ taking the limit as $zrightarrow fracpi2$, the second derivative gives you $2a_-1$ which is two times the residue in $fracpi2$
    – Davide Morgante
    Sep 3 at 14:02
















$pi / 2$ is a pole of order 3
– Davide Morgante
Sep 3 at 13:28




$pi / 2$ is a pole of order 3
– Davide Morgante
Sep 3 at 13:28












why ? (1-sinz) get the order 2 and the 2z-pi is order 1 so it should not be a pole
– xmaionx
Sep 3 at 13:30




why ? (1-sinz) get the order 2 and the 2z-pi is order 1 so it should not be a pole
– xmaionx
Sep 3 at 13:30












$operatornameord_z_0(fg) = operatornameord_z_0(f)+operatornameord_z_0(g)$
– Davide Morgante
Sep 3 at 13:32




$operatornameord_z_0(fg) = operatornameord_z_0(f)+operatornameord_z_0(g)$
– Davide Morgante
Sep 3 at 13:32












ah but how i can solve the tan part ?
– xmaionx
Sep 3 at 13:37





ah but how i can solve the tan part ?
– xmaionx
Sep 3 at 13:37













You can evaluate residues of higher order by derivation: e.g. $z=pi/2$ is a order 3 pole. The Laurent expansion of $$f(z) = fraca_-3(z-fracpi2)^3+ fraca_-2(z-fracpi2)^2+ fraca_-1(z-fracpi2)+a_0+cdots$$ so $$fracd^2dz^2(f(z)(z-fracpi2)^3)= fracd^2dz^2left(a_-3+(z-fracpi2)a_-2+(z-fracpi2)^2a_-1+cdotsright)$$ taking the limit as $zrightarrow fracpi2$, the second derivative gives you $2a_-1$ which is two times the residue in $fracpi2$
– Davide Morgante
Sep 3 at 14:02





You can evaluate residues of higher order by derivation: e.g. $z=pi/2$ is a order 3 pole. The Laurent expansion of $$f(z) = fraca_-3(z-fracpi2)^3+ fraca_-2(z-fracpi2)^2+ fraca_-1(z-fracpi2)+a_0+cdots$$ so $$fracd^2dz^2(f(z)(z-fracpi2)^3)= fracd^2dz^2left(a_-3+(z-fracpi2)a_-2+(z-fracpi2)^2a_-1+cdotsright)$$ taking the limit as $zrightarrow fracpi2$, the second derivative gives you $2a_-1$ which is two times the residue in $fracpi2$
– Davide Morgante
Sep 3 at 14:02
















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