Improper Integral Residue Theorem












0












$begingroup$


I'm stuck on a question involving evaluating improper integrals using the residue theorem.

Here's what I'm trying to evaluate:
$int_{-infty}^{infty} frac{1}{(1+x^2)^3} dx$



To begin with, we can define a contour integral over the the semi-circle disc above the Real axis called $C$ which can be broken up in two parts involving the Real number line, and the curve of the semi-circular disc
$int_{C} frac{1}{(1+z^2)^3} dz = int_{-R}^{R} frac{1}{(1+x^2)^3} dx + int_{Cr} frac{1}{(1+z^2)^3} dz$



We can see that the left-hand side of the equation has poles at $i$ and $-i$ with multiplicity $3$ respectively. Using the Residue Theorem
$int_{C} frac{1}{(1+z^2)^3} dz = 2 pi i (3 frac{1}{((i)+i)^3})$



Because there are $3$ instances of the $i$ being the pole inside the region defined.

Also:
$int_{Cr} frac{1}{(1+z^2)^3} dz = 0$

Because parameterizing $z = Re^{it}$ and setting the limit as R goes to infinity will show the integrand to be zero.



That leaves us with the final answer of
$-frac{6 pi}{8}$



But this is not correct.

The correct answer is $frac{3 pi}{8}$ and I'm not quite sure how that is formed.



Any help would be appreciated.

Thanks.






complex-analysis indefinite-integrals residue-calculus







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I'm stuck on a question involving evaluating improper integrals using the residue theorem.

    Here's what I'm trying to evaluate:
    $int_{-infty}^{infty} frac{1}{(1+x^2)^3} dx$



    To begin with, we can define a contour integral over the the semi-circle disc above the Real axis called $C$ which can be broken up in two parts involving the Real number line, and the curve of the semi-circular disc
    $int_{C} frac{1}{(1+z^2)^3} dz = int_{-R}^{R} frac{1}{(1+x^2)^3} dx + int_{Cr} frac{1}{(1+z^2)^3} dz$



    We can see that the left-hand side of the equation has poles at $i$ and $-i$ with multiplicity $3$ respectively. Using the Residue Theorem
    $int_{C} frac{1}{(1+z^2)^3} dz = 2 pi i (3 frac{1}{((i)+i)^3})$



    Because there are $3$ instances of the $i$ being the pole inside the region defined.

    Also:
    $int_{Cr} frac{1}{(1+z^2)^3} dz = 0$

    Because parameterizing $z = Re^{it}$ and setting the limit as R goes to infinity will show the integrand to be zero.



    That leaves us with the final answer of
    $-frac{6 pi}{8}$



    But this is not correct.

    The correct answer is $frac{3 pi}{8}$ and I'm not quite sure how that is formed.



    Any help would be appreciated.

    Thanks.






    complex-analysis indefinite-integrals residue-calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm stuck on a question involving evaluating improper integrals using the residue theorem.

      Here's what I'm trying to evaluate:
      $int_{-infty}^{infty} frac{1}{(1+x^2)^3} dx$



      To begin with, we can define a contour integral over the the semi-circle disc above the Real axis called $C$ which can be broken up in two parts involving the Real number line, and the curve of the semi-circular disc
      $int_{C} frac{1}{(1+z^2)^3} dz = int_{-R}^{R} frac{1}{(1+x^2)^3} dx + int_{Cr} frac{1}{(1+z^2)^3} dz$



      We can see that the left-hand side of the equation has poles at $i$ and $-i$ with multiplicity $3$ respectively. Using the Residue Theorem
      $int_{C} frac{1}{(1+z^2)^3} dz = 2 pi i (3 frac{1}{((i)+i)^3})$



      Because there are $3$ instances of the $i$ being the pole inside the region defined.

      Also:
      $int_{Cr} frac{1}{(1+z^2)^3} dz = 0$

      Because parameterizing $z = Re^{it}$ and setting the limit as R goes to infinity will show the integrand to be zero.



      That leaves us with the final answer of
      $-frac{6 pi}{8}$



      But this is not correct.

      The correct answer is $frac{3 pi}{8}$ and I'm not quite sure how that is formed.



      Any help would be appreciated.

      Thanks.






      complex-analysis indefinite-integrals residue-calculus







      share|cite|improve this question









      $endgroup$




      I'm stuck on a question involving evaluating improper integrals using the residue theorem.

      Here's what I'm trying to evaluate:
      $int_{-infty}^{infty} frac{1}{(1+x^2)^3} dx$



      To begin with, we can define a contour integral over the the semi-circle disc above the Real axis called $C$ which can be broken up in two parts involving the Real number line, and the curve of the semi-circular disc
      $int_{C} frac{1}{(1+z^2)^3} dz = int_{-R}^{R} frac{1}{(1+x^2)^3} dx + int_{Cr} frac{1}{(1+z^2)^3} dz$



      We can see that the left-hand side of the equation has poles at $i$ and $-i$ with multiplicity $3$ respectively. Using the Residue Theorem
      $int_{C} frac{1}{(1+z^2)^3} dz = 2 pi i (3 frac{1}{((i)+i)^3})$



      Because there are $3$ instances of the $i$ being the pole inside the region defined.

      Also:
      $int_{Cr} frac{1}{(1+z^2)^3} dz = 0$

      Because parameterizing $z = Re^{it}$ and setting the limit as R goes to infinity will show the integrand to be zero.



      That leaves us with the final answer of
      $-frac{6 pi}{8}$



      But this is not correct.

      The correct answer is $frac{3 pi}{8}$ and I'm not quite sure how that is formed.



      Any help would be appreciated.

      Thanks.






      complex-analysis indefinite-integrals residue-calculus




      complex-analysis indefinite-integrals residue-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 17:01









      user2965071user2965071

      1456




      1456






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The residue of $dfrac1{(1+z^2)^3}$ at $i$ is $-dfrac{3i}{16}$ and therefore your integral is equal to$$2pi itimesfrac{-3i}{16}=frac{3pi}8$$indeed.



          In order to compute that residue, you can use the fact that$$dfrac1{(1+z^2)^3}=frac{frac1{(z+i)^3}}{(z-i)^3}$$and that therefore the residue that we are interested in is $dfrac{varphi''(i)}2$, where $varphi(z)=dfrac1{(z+i)^3}$. And, as I wrote, $dfrac{varphi''(i)}2=-dfrac{3i}{16}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually?
            $endgroup$
            – user2965071
            Dec 13 '18 at 17:39










          • $begingroup$
            Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 17:51












          • $begingroup$
            Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial?
            $endgroup$
            – user2965071
            Dec 13 '18 at 17:56










          • $begingroup$
            I would not put it like that, but that's the right idea.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 18:08



















          0












          $begingroup$

          By the Cauchy integral formula



          if $a$ is inside the contour, and $f(z)$ is holomorphic inside the contour.



          $int_gamma frac {f(z)}{z-a} dz = 2pi i f(a)$



          and



          $int_gamma frac {f(z)}{(z-a)^n} dz = 2pi i frac {1}{(n-1)!} f^{(n-1)}(a)$



          in this case let $a = i$



          and $f(z) = frac {1}{(z+i)^3}$






          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%2f3038289%2fimproper-integral-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









            1












            $begingroup$

            The residue of $dfrac1{(1+z^2)^3}$ at $i$ is $-dfrac{3i}{16}$ and therefore your integral is equal to$$2pi itimesfrac{-3i}{16}=frac{3pi}8$$indeed.



            In order to compute that residue, you can use the fact that$$dfrac1{(1+z^2)^3}=frac{frac1{(z+i)^3}}{(z-i)^3}$$and that therefore the residue that we are interested in is $dfrac{varphi''(i)}2$, where $varphi(z)=dfrac1{(z+i)^3}$. And, as I wrote, $dfrac{varphi''(i)}2=-dfrac{3i}{16}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:39










            • $begingroup$
              Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 17:51












            • $begingroup$
              Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:56










            • $begingroup$
              I would not put it like that, but that's the right idea.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 18:08
















            1












            $begingroup$

            The residue of $dfrac1{(1+z^2)^3}$ at $i$ is $-dfrac{3i}{16}$ and therefore your integral is equal to$$2pi itimesfrac{-3i}{16}=frac{3pi}8$$indeed.



            In order to compute that residue, you can use the fact that$$dfrac1{(1+z^2)^3}=frac{frac1{(z+i)^3}}{(z-i)^3}$$and that therefore the residue that we are interested in is $dfrac{varphi''(i)}2$, where $varphi(z)=dfrac1{(z+i)^3}$. And, as I wrote, $dfrac{varphi''(i)}2=-dfrac{3i}{16}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:39










            • $begingroup$
              Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 17:51












            • $begingroup$
              Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:56










            • $begingroup$
              I would not put it like that, but that's the right idea.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 18:08














            1












            1








            1





            $begingroup$

            The residue of $dfrac1{(1+z^2)^3}$ at $i$ is $-dfrac{3i}{16}$ and therefore your integral is equal to$$2pi itimesfrac{-3i}{16}=frac{3pi}8$$indeed.



            In order to compute that residue, you can use the fact that$$dfrac1{(1+z^2)^3}=frac{frac1{(z+i)^3}}{(z-i)^3}$$and that therefore the residue that we are interested in is $dfrac{varphi''(i)}2$, where $varphi(z)=dfrac1{(z+i)^3}$. And, as I wrote, $dfrac{varphi''(i)}2=-dfrac{3i}{16}$.






            share|cite|improve this answer











            $endgroup$



            The residue of $dfrac1{(1+z^2)^3}$ at $i$ is $-dfrac{3i}{16}$ and therefore your integral is equal to$$2pi itimesfrac{-3i}{16}=frac{3pi}8$$indeed.



            In order to compute that residue, you can use the fact that$$dfrac1{(1+z^2)^3}=frac{frac1{(z+i)^3}}{(z-i)^3}$$and that therefore the residue that we are interested in is $dfrac{varphi''(i)}2$, where $varphi(z)=dfrac1{(z+i)^3}$. And, as I wrote, $dfrac{varphi''(i)}2=-dfrac{3i}{16}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 13 '18 at 17:13

























            answered Dec 13 '18 at 17:06









            José Carlos SantosJosé Carlos Santos

            164k22132235




            164k22132235












            • $begingroup$
              Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:39










            • $begingroup$
              Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 17:51












            • $begingroup$
              Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:56










            • $begingroup$
              I would not put it like that, but that's the right idea.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 18:08


















            • $begingroup$
              Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:39










            • $begingroup$
              Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 17:51












            • $begingroup$
              Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial?
              $endgroup$
              – user2965071
              Dec 13 '18 at 17:56










            • $begingroup$
              I would not put it like that, but that's the right idea.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 18:08
















            $begingroup$
            Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually?
            $endgroup$
            – user2965071
            Dec 13 '18 at 17:39




            $begingroup$
            Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually?
            $endgroup$
            – user2965071
            Dec 13 '18 at 17:39












            $begingroup$
            Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 17:51






            $begingroup$
            Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 17:51














            $begingroup$
            Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial?
            $endgroup$
            – user2965071
            Dec 13 '18 at 17:56




            $begingroup$
            Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial?
            $endgroup$
            – user2965071
            Dec 13 '18 at 17:56












            $begingroup$
            I would not put it like that, but that's the right idea.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 18:08




            $begingroup$
            I would not put it like that, but that's the right idea.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 18:08











            0












            $begingroup$

            By the Cauchy integral formula



            if $a$ is inside the contour, and $f(z)$ is holomorphic inside the contour.



            $int_gamma frac {f(z)}{z-a} dz = 2pi i f(a)$



            and



            $int_gamma frac {f(z)}{(z-a)^n} dz = 2pi i frac {1}{(n-1)!} f^{(n-1)}(a)$



            in this case let $a = i$



            and $f(z) = frac {1}{(z+i)^3}$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              By the Cauchy integral formula



              if $a$ is inside the contour, and $f(z)$ is holomorphic inside the contour.



              $int_gamma frac {f(z)}{z-a} dz = 2pi i f(a)$



              and



              $int_gamma frac {f(z)}{(z-a)^n} dz = 2pi i frac {1}{(n-1)!} f^{(n-1)}(a)$



              in this case let $a = i$



              and $f(z) = frac {1}{(z+i)^3}$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                By the Cauchy integral formula



                if $a$ is inside the contour, and $f(z)$ is holomorphic inside the contour.



                $int_gamma frac {f(z)}{z-a} dz = 2pi i f(a)$



                and



                $int_gamma frac {f(z)}{(z-a)^n} dz = 2pi i frac {1}{(n-1)!} f^{(n-1)}(a)$



                in this case let $a = i$



                and $f(z) = frac {1}{(z+i)^3}$






                share|cite|improve this answer









                $endgroup$



                By the Cauchy integral formula



                if $a$ is inside the contour, and $f(z)$ is holomorphic inside the contour.



                $int_gamma frac {f(z)}{z-a} dz = 2pi i f(a)$



                and



                $int_gamma frac {f(z)}{(z-a)^n} dz = 2pi i frac {1}{(n-1)!} f^{(n-1)}(a)$



                in this case let $a = i$



                and $f(z) = frac {1}{(z+i)^3}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 17:18









                Doug MDoug M

                45.3k31954




                45.3k31954






























                    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%2f3038289%2fimproper-integral-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

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                    Read more

                    Brian Clough

                    Image
                    Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
                    Read more

                    Cáceres

                    Image
                    Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
                    Read more