$lim_{rtoinfty}int_{c-ir}^{c+ir}z^{-lambda}(log(z))^{-1}dz$











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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?










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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?










    share|cite|improve this question
























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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?










      share|cite|improve this question













      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






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      asked Nov 26 at 1:55









      mathstackuser

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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 $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-lambda}Big/sqrt{t^2+log^2rho}leqslantrho^{-lambda}/logrho$, so that the integral over $gamma$ is at most
          $$frac{2phirho^{1-lambda}}{logrho}<frac{pirho^{1-lambda}}{logrho}<frac{pi r^{1-lambda}}{log r}.$$






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            up vote
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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 $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-lambda}Big/sqrt{t^2+log^2rho}leqslantrho^{-lambda}/logrho$, so that the integral over $gamma$ is at most
            $$frac{2phirho^{1-lambda}}{logrho}<frac{pirho^{1-lambda}}{logrho}<frac{pi r^{1-lambda}}{log r}.$$






            share|cite|improve this answer



























              up vote
              0
              down vote













              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 $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-lambda}Big/sqrt{t^2+log^2rho}leqslantrho^{-lambda}/logrho$, so that the integral over $gamma$ is at most
              $$frac{2phirho^{1-lambda}}{logrho}<frac{pirho^{1-lambda}}{logrho}<frac{pi r^{1-lambda}}{log r}.$$






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                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 $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-lambda}Big/sqrt{t^2+log^2rho}leqslantrho^{-lambda}/logrho$, so that the integral over $gamma$ is at most
                $$frac{2phirho^{1-lambda}}{logrho}<frac{pirho^{1-lambda}}{logrho}<frac{pi r^{1-lambda}}{log r}.$$






                share|cite|improve this answer














                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 $z=rho e^{it} implies |z^{-lambda}log^{-1}z|=rho^{-lambda}Big/sqrt{t^2+log^2rho}leqslantrho^{-lambda}/logrho$, so that the integral over $gamma$ is at most
                $$frac{2phirho^{1-lambda}}{logrho}<frac{pirho^{1-lambda}}{logrho}<frac{pi r^{1-lambda}}{log r}.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 1 at 22:39

























                answered Dec 1 at 22:34









                metamorphy

                2,9371517




                2,9371517






























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