Evaluating a definite integral using residue theorem












0












$begingroup$


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











$endgroup$

















    0












    $begingroup$


    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











    $endgroup$















      0












      0








      0





      $begingroup$


      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 10:06









      Tianlalu

      3,08621038




      3,08621038










      asked Dec 4 '18 at 1:13









      Richard VillalobosRichard Villalobos

      1567




      1567






















          2 Answers
          2






          active

          oldest

          votes


















          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



















          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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024977%2fevaluating-a-definite-integral-using-residue-theorem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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

                288k33280660




                288k33280660






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024977%2fevaluating-a-definite-integral-using-residue-theorem%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                    Read more

                    build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                    Image
                    0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                    Read more