Evaluating a definite integral using residue theorem












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I am trying to show $$int_0^{infty} frac{log (x)}{(x^2+1)^2}$$. We can use the integrand $g(z) = frac{log(z)}{(x^2+1)^2}$ defining log as $log(rho e^{itheta}) = log(rho) + itheta$ and letting $theta in [0,2pi]$



Now, I know that the singularities of the integrand occur at $i,-i$ one being in the upper half plane, the other in the lower half plane.



Typically the counter used for integrands which require choosing a branch cut of log is a half donut shape on the upper plane. Since there is a singularity no on the lower half plane, do I ignore it or include the residue of it in the calculation?



Also, I believe the singularities to be poles of order 3 and I do not know how to calculate residues for them. Is there an explicit formula like there is for simple poles?






complex-analysis residue-calculus







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    $begingroup$


    I am trying to show $$int_0^{infty} frac{log (x)}{(x^2+1)^2}$$. We can use the integrand $g(z) = frac{log(z)}{(x^2+1)^2}$ defining log as $log(rho e^{itheta}) = log(rho) + itheta$ and letting $theta in [0,2pi]$



    Now, I know that the singularities of the integrand occur at $i,-i$ one being in the upper half plane, the other in the lower half plane.



    Typically the counter used for integrands which require choosing a branch cut of log is a half donut shape on the upper plane. Since there is a singularity no on the lower half plane, do I ignore it or include the residue of it in the calculation?



    Also, I believe the singularities to be poles of order 3 and I do not know how to calculate residues for them. Is there an explicit formula like there is for simple poles?






    complex-analysis residue-calculus







    share|cite|improve this question











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      $begingroup$


      I am trying to show $$int_0^{infty} frac{log (x)}{(x^2+1)^2}$$. We can use the integrand $g(z) = frac{log(z)}{(x^2+1)^2}$ defining log as $log(rho e^{itheta}) = log(rho) + itheta$ and letting $theta in [0,2pi]$



      Now, I know that the singularities of the integrand occur at $i,-i$ one being in the upper half plane, the other in the lower half plane.



      Typically the counter used for integrands which require choosing a branch cut of log is a half donut shape on the upper plane. Since there is a singularity no on the lower half plane, do I ignore it or include the residue of it in the calculation?



      Also, I believe the singularities to be poles of order 3 and I do not know how to calculate residues for them. Is there an explicit formula like there is for simple poles?






      complex-analysis residue-calculus







      share|cite|improve this question











      $endgroup$




      I am trying to show $$int_0^{infty} frac{log (x)}{(x^2+1)^2}$$. We can use the integrand $g(z) = frac{log(z)}{(x^2+1)^2}$ defining log as $log(rho e^{itheta}) = log(rho) + itheta$ and letting $theta in [0,2pi]$



      Now, I know that the singularities of the integrand occur at $i,-i$ one being in the upper half plane, the other in the lower half plane.



      Typically the counter used for integrands which require choosing a branch cut of log is a half donut shape on the upper plane. Since there is a singularity no on the lower half plane, do I ignore it or include the residue of it in the calculation?



      Also, I believe the singularities to be poles of order 3 and I do not know how to calculate residues for them. Is there an explicit formula like there is for simple poles?






      complex-analysis residue-calculus




      complex-analysis residue-calculus






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      edited Dec 4 '18 at 10:06









      Tianlalu

      3,08621038




      3,08621038










      asked Dec 4 '18 at 1:13









      Richard VillalobosRichard Villalobos

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          $begingroup$

          Any singularity outside the contour will be irrelevant to the integral.



          The poles in this case will be of order $2$. A residue for a pole of order at most $m$ can be found from
          $$Res(f,z_0)=frac1{(m-1)!}lim_{zto z_0}frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)$$
          which for large $m$ can be frustrating but should be OK in this case.






          share|cite|improve this answer









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          • $begingroup$
            Ah yes, you are right that it is a pole of order 2, thank you for catching my mistake!
            $endgroup$
            – Richard Villalobos
            Dec 4 '18 at 1:25










          • $begingroup$
            so would I have $$Res(g,z_0) = lim_{z rightarrow i} 2(z-z_0) [frac{(z^2+1)^2}{z}-log(z)4z(z^2+1)]frac{1}{(z^2+1)^3}$$ and if so wouldnt this limit be zero?
            $endgroup$
            – Richard Villalobos
            Dec 4 '18 at 2:05












          • $begingroup$
            I can't follow your differentiation. Remember that in this case $z_0=i$, and that $$(z-z_0)^2f(z)=frac{(z-i)^2}{(z^2+1)^2}log z ,$$ simplify first then differentiate. The answer might be zero, I haven't worked it out.
            $endgroup$
            – David
            Dec 4 '18 at 3:00



















          1












          $begingroup$

          You are trying to show... what?
          $$begin{eqnarray*}int_{0}^{+infty}frac{log(x)}{(1+x^2)^2},dx &=& int_{0}^{1}frac{log x}{(1+x^2)^2},dx+underbrace{int_{0}^{1}frac{-x^2log(x)}{(1+x^2)^2},dx}_{xmapsto 1/x}\&=&int_{0}^{1}frac{1-x^2}{(1+x^2)^2}log(x),dxend{eqnarray*} $$
          and since $$forall xin(0,1),qquad frac{1-x^2}{(1+x^2)^2}=sum_{ngeq 0}(-1)^n (2n+1) x^{2n}, $$
          $$ int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$$
          we have:
          $$ int_{0}^{+infty}frac{log x}{(1+x^2)^2},dx = -sum_{ngeq 0}frac{(-1)^n}{2n+1}=-arctan(1)=color{red}{-frac{pi}{4}}.$$
          We do not strictly need Complex Analysis.






          share|cite|improve this answer









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            2 Answers
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            0












            $begingroup$

            Any singularity outside the contour will be irrelevant to the integral.



            The poles in this case will be of order $2$. A residue for a pole of order at most $m$ can be found from
            $$Res(f,z_0)=frac1{(m-1)!}lim_{zto z_0}frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)$$
            which for large $m$ can be frustrating but should be OK in this case.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ah yes, you are right that it is a pole of order 2, thank you for catching my mistake!
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 1:25










            • $begingroup$
              so would I have $$Res(g,z_0) = lim_{z rightarrow i} 2(z-z_0) [frac{(z^2+1)^2}{z}-log(z)4z(z^2+1)]frac{1}{(z^2+1)^3}$$ and if so wouldnt this limit be zero?
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 2:05












            • $begingroup$
              I can't follow your differentiation. Remember that in this case $z_0=i$, and that $$(z-z_0)^2f(z)=frac{(z-i)^2}{(z^2+1)^2}log z ,$$ simplify first then differentiate. The answer might be zero, I haven't worked it out.
              $endgroup$
              – David
              Dec 4 '18 at 3:00
















            0












            $begingroup$

            Any singularity outside the contour will be irrelevant to the integral.



            The poles in this case will be of order $2$. A residue for a pole of order at most $m$ can be found from
            $$Res(f,z_0)=frac1{(m-1)!}lim_{zto z_0}frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)$$
            which for large $m$ can be frustrating but should be OK in this case.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ah yes, you are right that it is a pole of order 2, thank you for catching my mistake!
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 1:25










            • $begingroup$
              so would I have $$Res(g,z_0) = lim_{z rightarrow i} 2(z-z_0) [frac{(z^2+1)^2}{z}-log(z)4z(z^2+1)]frac{1}{(z^2+1)^3}$$ and if so wouldnt this limit be zero?
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 2:05












            • $begingroup$
              I can't follow your differentiation. Remember that in this case $z_0=i$, and that $$(z-z_0)^2f(z)=frac{(z-i)^2}{(z^2+1)^2}log z ,$$ simplify first then differentiate. The answer might be zero, I haven't worked it out.
              $endgroup$
              – David
              Dec 4 '18 at 3:00














            0












            0








            0





            $begingroup$

            Any singularity outside the contour will be irrelevant to the integral.



            The poles in this case will be of order $2$. A residue for a pole of order at most $m$ can be found from
            $$Res(f,z_0)=frac1{(m-1)!}lim_{zto z_0}frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)$$
            which for large $m$ can be frustrating but should be OK in this case.






            share|cite|improve this answer









            $endgroup$



            Any singularity outside the contour will be irrelevant to the integral.



            The poles in this case will be of order $2$. A residue for a pole of order at most $m$ can be found from
            $$Res(f,z_0)=frac1{(m-1)!}lim_{zto z_0}frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)$$
            which for large $m$ can be frustrating but should be OK in this case.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 1:21









            DavidDavid

            68.1k664126




            68.1k664126












            • $begingroup$
              Ah yes, you are right that it is a pole of order 2, thank you for catching my mistake!
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 1:25










            • $begingroup$
              so would I have $$Res(g,z_0) = lim_{z rightarrow i} 2(z-z_0) [frac{(z^2+1)^2}{z}-log(z)4z(z^2+1)]frac{1}{(z^2+1)^3}$$ and if so wouldnt this limit be zero?
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 2:05












            • $begingroup$
              I can't follow your differentiation. Remember that in this case $z_0=i$, and that $$(z-z_0)^2f(z)=frac{(z-i)^2}{(z^2+1)^2}log z ,$$ simplify first then differentiate. The answer might be zero, I haven't worked it out.
              $endgroup$
              – David
              Dec 4 '18 at 3:00


















            • $begingroup$
              Ah yes, you are right that it is a pole of order 2, thank you for catching my mistake!
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 1:25










            • $begingroup$
              so would I have $$Res(g,z_0) = lim_{z rightarrow i} 2(z-z_0) [frac{(z^2+1)^2}{z}-log(z)4z(z^2+1)]frac{1}{(z^2+1)^3}$$ and if so wouldnt this limit be zero?
              $endgroup$
              – Richard Villalobos
              Dec 4 '18 at 2:05












            • $begingroup$
              I can't follow your differentiation. Remember that in this case $z_0=i$, and that $$(z-z_0)^2f(z)=frac{(z-i)^2}{(z^2+1)^2}log z ,$$ simplify first then differentiate. The answer might be zero, I haven't worked it out.
              $endgroup$
              – David
              Dec 4 '18 at 3:00
















            $begingroup$
            Ah yes, you are right that it is a pole of order 2, thank you for catching my mistake!
            $endgroup$
            – Richard Villalobos
            Dec 4 '18 at 1:25




            $begingroup$
            Ah yes, you are right that it is a pole of order 2, thank you for catching my mistake!
            $endgroup$
            – Richard Villalobos
            Dec 4 '18 at 1:25












            $begingroup$
            so would I have $$Res(g,z_0) = lim_{z rightarrow i} 2(z-z_0) [frac{(z^2+1)^2}{z}-log(z)4z(z^2+1)]frac{1}{(z^2+1)^3}$$ and if so wouldnt this limit be zero?
            $endgroup$
            – Richard Villalobos
            Dec 4 '18 at 2:05






            $begingroup$
            so would I have $$Res(g,z_0) = lim_{z rightarrow i} 2(z-z_0) [frac{(z^2+1)^2}{z}-log(z)4z(z^2+1)]frac{1}{(z^2+1)^3}$$ and if so wouldnt this limit be zero?
            $endgroup$
            – Richard Villalobos
            Dec 4 '18 at 2:05














            $begingroup$
            I can't follow your differentiation. Remember that in this case $z_0=i$, and that $$(z-z_0)^2f(z)=frac{(z-i)^2}{(z^2+1)^2}log z ,$$ simplify first then differentiate. The answer might be zero, I haven't worked it out.
            $endgroup$
            – David
            Dec 4 '18 at 3:00




            $begingroup$
            I can't follow your differentiation. Remember that in this case $z_0=i$, and that $$(z-z_0)^2f(z)=frac{(z-i)^2}{(z^2+1)^2}log z ,$$ simplify first then differentiate. The answer might be zero, I haven't worked it out.
            $endgroup$
            – David
            Dec 4 '18 at 3:00











            1












            $begingroup$

            You are trying to show... what?
            $$begin{eqnarray*}int_{0}^{+infty}frac{log(x)}{(1+x^2)^2},dx &=& int_{0}^{1}frac{log x}{(1+x^2)^2},dx+underbrace{int_{0}^{1}frac{-x^2log(x)}{(1+x^2)^2},dx}_{xmapsto 1/x}\&=&int_{0}^{1}frac{1-x^2}{(1+x^2)^2}log(x),dxend{eqnarray*} $$
            and since $$forall xin(0,1),qquad frac{1-x^2}{(1+x^2)^2}=sum_{ngeq 0}(-1)^n (2n+1) x^{2n}, $$
            $$ int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$$
            we have:
            $$ int_{0}^{+infty}frac{log x}{(1+x^2)^2},dx = -sum_{ngeq 0}frac{(-1)^n}{2n+1}=-arctan(1)=color{red}{-frac{pi}{4}}.$$
            We do not strictly need Complex Analysis.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You are trying to show... what?
              $$begin{eqnarray*}int_{0}^{+infty}frac{log(x)}{(1+x^2)^2},dx &=& int_{0}^{1}frac{log x}{(1+x^2)^2},dx+underbrace{int_{0}^{1}frac{-x^2log(x)}{(1+x^2)^2},dx}_{xmapsto 1/x}\&=&int_{0}^{1}frac{1-x^2}{(1+x^2)^2}log(x),dxend{eqnarray*} $$
              and since $$forall xin(0,1),qquad frac{1-x^2}{(1+x^2)^2}=sum_{ngeq 0}(-1)^n (2n+1) x^{2n}, $$
              $$ int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$$
              we have:
              $$ int_{0}^{+infty}frac{log x}{(1+x^2)^2},dx = -sum_{ngeq 0}frac{(-1)^n}{2n+1}=-arctan(1)=color{red}{-frac{pi}{4}}.$$
              We do not strictly need Complex Analysis.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You are trying to show... what?
                $$begin{eqnarray*}int_{0}^{+infty}frac{log(x)}{(1+x^2)^2},dx &=& int_{0}^{1}frac{log x}{(1+x^2)^2},dx+underbrace{int_{0}^{1}frac{-x^2log(x)}{(1+x^2)^2},dx}_{xmapsto 1/x}\&=&int_{0}^{1}frac{1-x^2}{(1+x^2)^2}log(x),dxend{eqnarray*} $$
                and since $$forall xin(0,1),qquad frac{1-x^2}{(1+x^2)^2}=sum_{ngeq 0}(-1)^n (2n+1) x^{2n}, $$
                $$ int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$$
                we have:
                $$ int_{0}^{+infty}frac{log x}{(1+x^2)^2},dx = -sum_{ngeq 0}frac{(-1)^n}{2n+1}=-arctan(1)=color{red}{-frac{pi}{4}}.$$
                We do not strictly need Complex Analysis.






                share|cite|improve this answer









                $endgroup$



                You are trying to show... what?
                $$begin{eqnarray*}int_{0}^{+infty}frac{log(x)}{(1+x^2)^2},dx &=& int_{0}^{1}frac{log x}{(1+x^2)^2},dx+underbrace{int_{0}^{1}frac{-x^2log(x)}{(1+x^2)^2},dx}_{xmapsto 1/x}\&=&int_{0}^{1}frac{1-x^2}{(1+x^2)^2}log(x),dxend{eqnarray*} $$
                and since $$forall xin(0,1),qquad frac{1-x^2}{(1+x^2)^2}=sum_{ngeq 0}(-1)^n (2n+1) x^{2n}, $$
                $$ int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$$
                we have:
                $$ int_{0}^{+infty}frac{log x}{(1+x^2)^2},dx = -sum_{ngeq 0}frac{(-1)^n}{2n+1}=-arctan(1)=color{red}{-frac{pi}{4}}.$$
                We do not strictly need Complex Analysis.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 10:51









                Jack D'AurizioJack D'Aurizio

                1




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