Mixing
Loop integral around pole lying on logarithmic branch cut
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I want to evaluate a general case of an integral
$$lim_rto0^+oint_ f(z)ln(z-k)dz$$
where
$f$ is meromorphic on the whole complex plane and $q$ is a pole/essential singularity of $f$.
$k,qinmathbb C$ are constants, with $kne q$.
$q$ lies on the branch cut of $ln(z-k)$.
$operatornamearg(z-k)in [theta_0,theta_0+2pi]$
Of course, the integral does not go across the branch cut. It should be understood as
$$int_theta_0^+^(theta_0+pi)^-+int^(theta_0+2pi)^-_(theta_0+pi)^+$$
If we know the Laurent series of $f$ around $q$ (say, $f(z)=sum^infty_n=-inftya_n(z-q)^n$), can we express the integral purely in terms of the coefficients above?
As far as I know, the integral might be equal to
$$int^2pi_0 f(q+re^it)ln(q+re^it-k)ire^itdt+(2pi i)^2operatorname*Res_z=qf(z)$$
However, I am particularly uncomfortable with the argument ‘because the integral switched branch cut so add $2pi i$ after the $ln$’.
Moreover, this expression is not purely in terms of the Laurent coefficients.
Please evaluate $$lim_rto0^+oint_ f(z)ln(z-k)dz$$ in terms of $a_n$ where $f(z)=sum^infty_n=-inftya_n(z-q)^n$ in a neighborhood of $q$. Many thanks.
contour-integration residue-calculus
asked Jul 30 at 9:57
Szeto
3,8431421
add a comment |Â
up vote
0
down vote
favorite
I want to evaluate a general case of an integral
$$lim_rto0^+oint_ f(z)ln(z-k)dz$$
where
$f$ is meromorphic on the whole complex plane and $q$ is a pole/essential singularity of $f$.
$k,qinmathbb C$ are constants, with $kne q$.
$q$ lies on the branch cut of $ln(z-k)$.
$operatornamearg(z-k)in [theta_0,theta_0+2pi]$
Of course, the integral does not go across the branch cut. It should be understood as
$$int_theta_0^+^(theta_0+pi)^-+int^(theta_0+2pi)^-_(theta_0+pi)^+$$
If we know the Laurent series of $f$ around $q$ (say, $f(z)=sum^infty_n=-inftya_n(z-q)^n$), can we express the integral purely in terms of the coefficients above?
As far as I know, the integral might be equal to
$$int^2pi_0 f(q+re^it)ln(q+re^it-k)ire^itdt+(2pi i)^2operatorname*Res_z=qf(z)$$
However, I am particularly uncomfortable with the argument ‘because the integral switched branch cut so add $2pi i$ after the $ln$’.
Moreover, this expression is not purely in terms of the Laurent coefficients.
Please evaluate $$lim_rto0^+oint_ f(z)ln(z-k)dz$$ in terms of $a_n$ where $f(z)=sum^infty_n=-inftya_n(z-q)^n$ in a neighborhood of $q$. Many thanks.
contour-integration residue-calculus
asked Jul 30 at 9:57
Szeto
3,8431421
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to evaluate a general case of an integral
$$lim_rto0^+oint_ f(z)ln(z-k)dz$$
where
$f$ is meromorphic on the whole complex plane and $q$ is a pole/essential singularity of $f$.
$k,qinmathbb C$ are constants, with $kne q$.
$q$ lies on the branch cut of $ln(z-k)$.
$operatornamearg(z-k)in [theta_0,theta_0+2pi]$
Of course, the integral does not go across the branch cut. It should be understood as
$$int_theta_0^+^(theta_0+pi)^-+int^(theta_0+2pi)^-_(theta_0+pi)^+$$
If we know the Laurent series of $f$ around $q$ (say, $f(z)=sum^infty_n=-inftya_n(z-q)^n$), can we express the integral purely in terms of the coefficients above?
As far as I know, the integral might be equal to
$$int^2pi_0 f(q+re^it)ln(q+re^it-k)ire^itdt+(2pi i)^2operatorname*Res_z=qf(z)$$
However, I am particularly uncomfortable with the argument ‘because the integral switched branch cut so add $2pi i$ after the $ln$’.
Moreover, this expression is not purely in terms of the Laurent coefficients.
Please evaluate $$lim_rto0^+oint_ f(z)ln(z-k)dz$$ in terms of $a_n$ where $f(z)=sum^infty_n=-inftya_n(z-q)^n$ in a neighborhood of $q$. Many thanks.
contour-integration residue-calculus
asked Jul 30 at 9:57
Szeto
3,8431421
I want to evaluate a general case of an integral
$$lim_rto0^+oint_ f(z)ln(z-k)dz$$
where
$f$ is meromorphic on the whole complex plane and $q$ is a pole/essential singularity of $f$.
$k,qinmathbb C$ are constants, with $kne q$.
$q$ lies on the branch cut of $ln(z-k)$.
$operatornamearg(z-k)in [theta_0,theta_0+2pi]$
Of course, the integral does not go across the branch cut. It should be understood as
$$int_theta_0^+^(theta_0+pi)^-+int^(theta_0+2pi)^-_(theta_0+pi)^+$$
If we know the Laurent series of $f$ around $q$ (say, $f(z)=sum^infty_n=-inftya_n(z-q)^n$), can we express the integral purely in terms of the coefficients above?
As far as I know, the integral might be equal to
$$int^2pi_0 f(q+re^it)ln(q+re^it-k)ire^itdt+(2pi i)^2operatorname*Res_z=qf(z)$$
However, I am particularly uncomfortable with the argument ‘because the integral switched branch cut so add $2pi i$ after the $ln$’.
Moreover, this expression is not purely in terms of the Laurent coefficients.
Please evaluate $$lim_rto0^+oint_ f(z)ln(z-k)dz$$ in terms of $a_n$ where $f(z)=sum^infty_n=-inftya_n(z-q)^n$ in a neighborhood of $q$. Many thanks.
contour-integration residue-calculus
asked Jul 30 at 9:57
Szeto
3,8431421
edited Jul 30 at 11:25
asked Jul 30 at 9:57
Szeto
3,8431421
asked Jul 30 at 9:57
Szeto
3,8431421
asked Jul 30 at 9:57
Szeto
3,8431421
3,8431421
add a comment |Â
add a comment |Â
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