A line integral involving $zeta(s)$












1












$begingroup$


Consider the line integral



$$I=int_{1/2 -iinfty}^{1/2 + iinfty} frac{log((s-1)zeta(s))}{s} mathrm{d}s-int_{1/2 -i infty}^{1/2 + iinfty} frac{iarg zeta (s)}{s}mathrm{d}s$$



where $zeta$ denotes the Riemann zeta function, using complex analysis (or otherwise). Note that $I$ exists (is well-defined) since $|zeta(s)|=o(|s|)$ for $Re(s)=1/2$.



Is the argument of https://arxiv.org/abs/1306.0856 (proofs of Theorem 1.2-1.3) applicable in evaluating $I$ ?






complex-analysis analytic-number-theory riemann-zeta







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$endgroup$

















    1












    $begingroup$


    Consider the line integral



    $$I=int_{1/2 -iinfty}^{1/2 + iinfty} frac{log((s-1)zeta(s))}{s} mathrm{d}s-int_{1/2 -i infty}^{1/2 + iinfty} frac{iarg zeta (s)}{s}mathrm{d}s$$



    where $zeta$ denotes the Riemann zeta function, using complex analysis (or otherwise). Note that $I$ exists (is well-defined) since $|zeta(s)|=o(|s|)$ for $Re(s)=1/2$.



    Is the argument of https://arxiv.org/abs/1306.0856 (proofs of Theorem 1.2-1.3) applicable in evaluating $I$ ?






    complex-analysis analytic-number-theory riemann-zeta







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Consider the line integral



      $$I=int_{1/2 -iinfty}^{1/2 + iinfty} frac{log((s-1)zeta(s))}{s} mathrm{d}s-int_{1/2 -i infty}^{1/2 + iinfty} frac{iarg zeta (s)}{s}mathrm{d}s$$



      where $zeta$ denotes the Riemann zeta function, using complex analysis (or otherwise). Note that $I$ exists (is well-defined) since $|zeta(s)|=o(|s|)$ for $Re(s)=1/2$.



      Is the argument of https://arxiv.org/abs/1306.0856 (proofs of Theorem 1.2-1.3) applicable in evaluating $I$ ?






      complex-analysis analytic-number-theory riemann-zeta







      share|cite|improve this question











      $endgroup$




      Consider the line integral



      $$I=int_{1/2 -iinfty}^{1/2 + iinfty} frac{log((s-1)zeta(s))}{s} mathrm{d}s-int_{1/2 -i infty}^{1/2 + iinfty} frac{iarg zeta (s)}{s}mathrm{d}s$$



      where $zeta$ denotes the Riemann zeta function, using complex analysis (or otherwise). Note that $I$ exists (is well-defined) since $|zeta(s)|=o(|s|)$ for $Re(s)=1/2$.



      Is the argument of https://arxiv.org/abs/1306.0856 (proofs of Theorem 1.2-1.3) applicable in evaluating $I$ ?






      complex-analysis analytic-number-theory riemann-zeta




      complex-analysis analytic-number-theory riemann-zeta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 4 '18 at 2:07

























      asked Dec 3 '18 at 20:21







      user507152





























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          • Do you know the Cauchy integral theorem ? $$int_{1/2 -iinfty}^{1/2 + iinfty} frac{log(s-1)}{s^2} ds= lim_{ato 0}int_{C_a} frac{log(s-1)}{s^2} ds - int_{L_a} frac{log(s-1)}{s^2} ds$$
            where $C_a$ is a closed-contour $1/2-iinfty to 1/2-ia to 1+a-ia to 1+a+iato 1/2+ia to 1/2+iinfty to infty +iinfty to infty-iinfty to 1/2-iinfty$ and $L_a$ is the portion $1/2-ia to 1+a-ia to 1+a+iato 1/2+ia$.


          Since $frac{log(s-1)}{s^2}$ is analytic inside $S_a$ and decreases in $O(s^{-1-epsilon})$ then $int_{C_a} frac{log(s-1)}{s^2} ds = 0$ and since $log(s-1) mapsto log(s-1)+2ipi$ when rotating counterclockwise around $s-1$ then
          $$lim_{ato 0}int_{L_a} frac{log(s-1)}{s^2} ds= int_1^{1/2} frac{2ipi}{s^2}ds$$




          • For $F$ analytic decreasing in $O(s^{-1-epsilon})$, assuming the implied branch of $log (zeta(s)(s-1))$ is $O(s^{c}), c < epsilon$ the same argument (with a contour excluding each non-trivial zero) gives
            $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} log(zeta(s)(s-1))F(s) ds = lim_{T to infty}sum_{rho = sigma+it, sigma > 1/2, |t| < T} int_{1/2+it}^{sigma+it} 2i pi F(s)ds$$



          • In your question, what branch of $log (zeta(s)(s-1))$ are you considering ?



            Look at $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} frac{log(zeta(s)(s-1))}{s^{1+1/T}} ds$$








          share|cite|improve this answer











          $endgroup$













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            $begingroup$


            • Do you know the Cauchy integral theorem ? $$int_{1/2 -iinfty}^{1/2 + iinfty} frac{log(s-1)}{s^2} ds= lim_{ato 0}int_{C_a} frac{log(s-1)}{s^2} ds - int_{L_a} frac{log(s-1)}{s^2} ds$$
              where $C_a$ is a closed-contour $1/2-iinfty to 1/2-ia to 1+a-ia to 1+a+iato 1/2+ia to 1/2+iinfty to infty +iinfty to infty-iinfty to 1/2-iinfty$ and $L_a$ is the portion $1/2-ia to 1+a-ia to 1+a+iato 1/2+ia$.


            Since $frac{log(s-1)}{s^2}$ is analytic inside $S_a$ and decreases in $O(s^{-1-epsilon})$ then $int_{C_a} frac{log(s-1)}{s^2} ds = 0$ and since $log(s-1) mapsto log(s-1)+2ipi$ when rotating counterclockwise around $s-1$ then
            $$lim_{ato 0}int_{L_a} frac{log(s-1)}{s^2} ds= int_1^{1/2} frac{2ipi}{s^2}ds$$




            • For $F$ analytic decreasing in $O(s^{-1-epsilon})$, assuming the implied branch of $log (zeta(s)(s-1))$ is $O(s^{c}), c < epsilon$ the same argument (with a contour excluding each non-trivial zero) gives
              $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} log(zeta(s)(s-1))F(s) ds = lim_{T to infty}sum_{rho = sigma+it, sigma > 1/2, |t| < T} int_{1/2+it}^{sigma+it} 2i pi F(s)ds$$



            • In your question, what branch of $log (zeta(s)(s-1))$ are you considering ?



              Look at $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} frac{log(zeta(s)(s-1))}{s^{1+1/T}} ds$$








            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$


              • Do you know the Cauchy integral theorem ? $$int_{1/2 -iinfty}^{1/2 + iinfty} frac{log(s-1)}{s^2} ds= lim_{ato 0}int_{C_a} frac{log(s-1)}{s^2} ds - int_{L_a} frac{log(s-1)}{s^2} ds$$
                where $C_a$ is a closed-contour $1/2-iinfty to 1/2-ia to 1+a-ia to 1+a+iato 1/2+ia to 1/2+iinfty to infty +iinfty to infty-iinfty to 1/2-iinfty$ and $L_a$ is the portion $1/2-ia to 1+a-ia to 1+a+iato 1/2+ia$.


              Since $frac{log(s-1)}{s^2}$ is analytic inside $S_a$ and decreases in $O(s^{-1-epsilon})$ then $int_{C_a} frac{log(s-1)}{s^2} ds = 0$ and since $log(s-1) mapsto log(s-1)+2ipi$ when rotating counterclockwise around $s-1$ then
              $$lim_{ato 0}int_{L_a} frac{log(s-1)}{s^2} ds= int_1^{1/2} frac{2ipi}{s^2}ds$$




              • For $F$ analytic decreasing in $O(s^{-1-epsilon})$, assuming the implied branch of $log (zeta(s)(s-1))$ is $O(s^{c}), c < epsilon$ the same argument (with a contour excluding each non-trivial zero) gives
                $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} log(zeta(s)(s-1))F(s) ds = lim_{T to infty}sum_{rho = sigma+it, sigma > 1/2, |t| < T} int_{1/2+it}^{sigma+it} 2i pi F(s)ds$$



              • In your question, what branch of $log (zeta(s)(s-1))$ are you considering ?



                Look at $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} frac{log(zeta(s)(s-1))}{s^{1+1/T}} ds$$








              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$


                • Do you know the Cauchy integral theorem ? $$int_{1/2 -iinfty}^{1/2 + iinfty} frac{log(s-1)}{s^2} ds= lim_{ato 0}int_{C_a} frac{log(s-1)}{s^2} ds - int_{L_a} frac{log(s-1)}{s^2} ds$$
                  where $C_a$ is a closed-contour $1/2-iinfty to 1/2-ia to 1+a-ia to 1+a+iato 1/2+ia to 1/2+iinfty to infty +iinfty to infty-iinfty to 1/2-iinfty$ and $L_a$ is the portion $1/2-ia to 1+a-ia to 1+a+iato 1/2+ia$.


                Since $frac{log(s-1)}{s^2}$ is analytic inside $S_a$ and decreases in $O(s^{-1-epsilon})$ then $int_{C_a} frac{log(s-1)}{s^2} ds = 0$ and since $log(s-1) mapsto log(s-1)+2ipi$ when rotating counterclockwise around $s-1$ then
                $$lim_{ato 0}int_{L_a} frac{log(s-1)}{s^2} ds= int_1^{1/2} frac{2ipi}{s^2}ds$$




                • For $F$ analytic decreasing in $O(s^{-1-epsilon})$, assuming the implied branch of $log (zeta(s)(s-1))$ is $O(s^{c}), c < epsilon$ the same argument (with a contour excluding each non-trivial zero) gives
                  $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} log(zeta(s)(s-1))F(s) ds = lim_{T to infty}sum_{rho = sigma+it, sigma > 1/2, |t| < T} int_{1/2+it}^{sigma+it} 2i pi F(s)ds$$



                • In your question, what branch of $log (zeta(s)(s-1))$ are you considering ?



                  Look at $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} frac{log(zeta(s)(s-1))}{s^{1+1/T}} ds$$








                share|cite|improve this answer











                $endgroup$




                • Do you know the Cauchy integral theorem ? $$int_{1/2 -iinfty}^{1/2 + iinfty} frac{log(s-1)}{s^2} ds= lim_{ato 0}int_{C_a} frac{log(s-1)}{s^2} ds - int_{L_a} frac{log(s-1)}{s^2} ds$$
                  where $C_a$ is a closed-contour $1/2-iinfty to 1/2-ia to 1+a-ia to 1+a+iato 1/2+ia to 1/2+iinfty to infty +iinfty to infty-iinfty to 1/2-iinfty$ and $L_a$ is the portion $1/2-ia to 1+a-ia to 1+a+iato 1/2+ia$.


                Since $frac{log(s-1)}{s^2}$ is analytic inside $S_a$ and decreases in $O(s^{-1-epsilon})$ then $int_{C_a} frac{log(s-1)}{s^2} ds = 0$ and since $log(s-1) mapsto log(s-1)+2ipi$ when rotating counterclockwise around $s-1$ then
                $$lim_{ato 0}int_{L_a} frac{log(s-1)}{s^2} ds= int_1^{1/2} frac{2ipi}{s^2}ds$$




                • For $F$ analytic decreasing in $O(s^{-1-epsilon})$, assuming the implied branch of $log (zeta(s)(s-1))$ is $O(s^{c}), c < epsilon$ the same argument (with a contour excluding each non-trivial zero) gives
                  $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} log(zeta(s)(s-1))F(s) ds = lim_{T to infty}sum_{rho = sigma+it, sigma > 1/2, |t| < T} int_{1/2+it}^{sigma+it} 2i pi F(s)ds$$



                • In your question, what branch of $log (zeta(s)(s-1))$ are you considering ?



                  Look at $$lim_{T to infty}int_{1/2 -iT}^{1/2 + iT} frac{log(zeta(s)(s-1))}{s^{1+1/T}} ds$$









                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 3 '18 at 22:47

























                answered Dec 3 '18 at 22:37









                reunsreuns

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                19.9k21148






























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