$lim_{rtoinfty}int_{c-ir}^{c+ir}z^{-lambda}(log(z))^{-1}dz$
Let $c>1$ and $lambdainmathbb{C}$ with $Re(lambda)geq1 $. Show that $lim_{rtoinfty}int_{c-ir}^{c+ir}z^{-lambda}(log(z))^{-1}dz=0$, where the path of integration is the straight line and $log$ denotes the principle branch of the logarithm.
I had the following idea: Closing the path with the semicircle $gamma(t)=c+re^{it}$ for $tin[-pi/2,pi/2]$. Then for a fixed $r$ the integral over this simple closed contour is $0$ by Cauchy's Theorem because the function is holomorphic inside and on the contour. So I am left to show that $lim_{rtoinfty}int_gamma z^{-lambda}(log(z))^{-1}dz=0$. I think at this point I have to use the definition of a path integral using the parametrization of $gamma$ and then estimate the absolut value of the integral hoping that I get something like $1/r$, but I have difficulties with estimating the absolut value of $dfrac{1}{e^{lambdalog(c+re^{it})}log(c+re^{it})}$. The derivative of $gamma$ just gives me a factor $r$ and then I also have to multiply by the length of the curve which is $pi r$.
Am I on the right way? Can anyone help me with the estimate?
integration complex-analysis limits logarithms estimation
asked Nov 26 at 1:55
mathstackuser
676112
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Let $c>1$ and $lambdainmathbb{C}$ with $Re(lambda)geq1 $. Show that $lim_{rtoinfty}int_{c-ir}^{c+ir}z^{-lambda}(log(z))^{-1}dz=0$, where the path of integration is the straight line and $log$ denotes the principle branch of the logarithm.
I had the following idea: Closing the path with the semicircle $gamma(t)=c+re^{it}$ for $tin[-pi/2,pi/2]$. Then for a fixed $r$ the integral over this simple closed contour is $0$ by Cauchy's Theorem because the function is holomorphic inside and on the contour. So I am left to show that $lim_{rtoinfty}int_gamma z^{-lambda}(log(z))^{-1}dz=0$. I think at this point I have to use the definition of a path integral using the parametrization of $gamma$ and then estimate the absolut value of the integral hoping that I get something like $1/r$, but I have difficulties with estimating the absolut value of $dfrac{1}{e^{lambdalog(c+re^{it})}log(c+re^{it})}$. The derivative of $gamma$ just gives me a factor $r$ and then I also have to multiply by the length of the curve which is $pi r$.
Am I on the right way? Can anyone help me with the estimate?
integration complex-analysis limits logarithms estimation
asked Nov 26 at 1:55
mathstackuser
676112
add a comment |
Let $c>1$ and $lambdainmathbb{C}$ with $Re(lambda)geq1 $. Show that $lim_{rtoinfty}int_{c-ir}^{c+ir}z^{-lambda}(log(z))^{-1}dz=0$, where the path of integration is the straight line and $log$ denotes the principle branch of the logarithm.
I had the following idea: Closing the path with the semicircle $gamma(t)=c+re^{it}$ for $tin[-pi/2,pi/2]$. Then for a fixed $r$ the integral over this simple closed contour is $0$ by Cauchy's Theorem because the function is holomorphic inside and on the contour. So I am left to show that $lim_{rtoinfty}int_gamma z^{-lambda}(log(z))^{-1}dz=0$. I think at this point I have to use the definition of a path integral using the parametrization of $gamma$ and then estimate the absolut value of the integral hoping that I get something like $1/r$, but I have difficulties with estimating the absolut value of $dfrac{1}{e^{lambdalog(c+re^{it})}log(c+re^{it})}$. The derivative of $gamma$ just gives me a factor $r$ and then I also have to multiply by the length of the curve which is $pi r$.
Am I on the right way? Can anyone help me with the estimate?
integration complex-analysis limits logarithms estimation
asked Nov 26 at 1:55
mathstackuser
676112
Let $c>1$ and $lambdainmathbb{C}$ with $Re(lambda)geq1 $. Show that $lim_{rtoinfty}int_{c-ir}^{c+ir}z^{-lambda}(log(z))^{-1}dz=0$, where the path of integration is the straight line and $log$ denotes the principle branch of the logarithm.
I had the following idea: Closing the path with the semicircle $gamma(t)=c+re^{it}$ for $tin[-pi/2,pi/2]$. Then for a fixed $r$ the integral over this simple closed contour is $0$ by Cauchy's Theorem because the function is holomorphic inside and on the contour. So I am left to show that $lim_{rtoinfty}int_gamma z^{-lambda}(log(z))^{-1}dz=0$. I think at this point I have to use the definition of a path integral using the parametrization of $gamma$ and then estimate the absolut value of the integral hoping that I get something like $1/r$, but I have difficulties with estimating the absolut value of $dfrac{1}{e^{lambdalog(c+re^{it})}log(c+re^{it})}$. The derivative of $gamma$ just gives me a factor $r$ and then I also have to multiply by the length of the curve which is $pi r$.
Am I on the right way? Can anyone help me with the estimate?
integration complex-analysis limits logarithms estimation
integration complex-analysis limits logarithms estimation
asked Nov 26 at 1:55
mathstackuser
676112
asked Nov 26 at 1:55
mathstackuser
676112
asked Nov 26 at 1:55
mathstackuser
676112
asked Nov 26 at 1:55
mathstackuser
676112
asked Nov 26 at 1:55
mathstackuser
676112
676112
add a comment |
add a comment |
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Yes, you are (just keep in mind that even a rough estimate will do).
You can go even a simpler way: let $c+ir=rho e^{iphi}$, choose $gamma(t)=rho e^{it}$ for $tin[-phi,phi]$ instead, and see that $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-sigma}Big/sqrt{t^2+log^2rho}leqslantrho^{-sigma}/logrho$ with $sigma=Relambda$, so that the integral over $gamma$ is at most
$$frac{2phirho^{1-sigma}}{logrho}<frac{pirho^{1-sigma}}{logrho}.$$
answered Dec 1 at 22:34
metamorphy
3,3671521
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1 Answer
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Yes, you are (just keep in mind that even a rough estimate will do).
You can go even a simpler way: let $c+ir=rho e^{iphi}$, choose $gamma(t)=rho e^{it}$ for $tin[-phi,phi]$ instead, and see that $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-sigma}Big/sqrt{t^2+log^2rho}leqslantrho^{-sigma}/logrho$ with $sigma=Relambda$, so that the integral over $gamma$ is at most
$$frac{2phirho^{1-sigma}}{logrho}<frac{pirho^{1-sigma}}{logrho}.$$
answered Dec 1 at 22:34
metamorphy
3,3671521
add a comment |
Yes, you are (just keep in mind that even a rough estimate will do).
You can go even a simpler way: let $c+ir=rho e^{iphi}$, choose $gamma(t)=rho e^{it}$ for $tin[-phi,phi]$ instead, and see that $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-sigma}Big/sqrt{t^2+log^2rho}leqslantrho^{-sigma}/logrho$ with $sigma=Relambda$, so that the integral over $gamma$ is at most
$$frac{2phirho^{1-sigma}}{logrho}<frac{pirho^{1-sigma}}{logrho}.$$
answered Dec 1 at 22:34
metamorphy
3,3671521
add a comment |
Yes, you are (just keep in mind that even a rough estimate will do).
You can go even a simpler way: let $c+ir=rho e^{iphi}$, choose $gamma(t)=rho e^{it}$ for $tin[-phi,phi]$ instead, and see that $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-sigma}Big/sqrt{t^2+log^2rho}leqslantrho^{-sigma}/logrho$ with $sigma=Relambda$, so that the integral over $gamma$ is at most
$$frac{2phirho^{1-sigma}}{logrho}<frac{pirho^{1-sigma}}{logrho}.$$
answered Dec 1 at 22:34
metamorphy
3,3671521
Yes, you are (just keep in mind that even a rough estimate will do).
You can go even a simpler way: let $c+ir=rho e^{iphi}$, choose $gamma(t)=rho e^{it}$ for $tin[-phi,phi]$ instead, and see that $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-sigma}Big/sqrt{t^2+log^2rho}leqslantrho^{-sigma}/logrho$ with $sigma=Relambda$, so that the integral over $gamma$ is at most
$$frac{2phirho^{1-sigma}}{logrho}<frac{pirho^{1-sigma}}{logrho}.$$
answered Dec 1 at 22:34
metamorphy
3,3671521
edited Dec 13 at 8:00
answered Dec 1 at 22:34
metamorphy
3,3671521
answered Dec 1 at 22:34
metamorphy
3,3671521
answered Dec 1 at 22:34
metamorphy
3,3671521
3,3671521
add a comment |
add a comment |
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