When to resolve into partial fractions for applying Cauchy's integral formula?












0












$begingroup$


Suppose I have to calculate $int frac{f(z)}{(z-a)(z-b)}dz$ around a curve in which both $a$ and $b$ lie inside.



Should I apply Residue theorem like this:



Residue at $a$= $frac{f(a)}{a-b}$



Residue at $b$= $frac{f(b)}{b-a}$



Then I use the Residue theorem 2pi*(sum of residues)



OR



Should I solve it by partial fractions?



Are both valid?



My confusion with the first method is that the numerator of $int frac{f(z)/(z-a)}{(z-b)}$, i.e. $frac{f(z)}{z-a}$ is not analytic, because $a$ also lies inside the curve. Aren't we allowed to use Cauchy's integral formula only when the numerator is analytic?






complex-analysis residue-calculus cauchy-integral-formula analytic-functions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Suppose I have to calculate $int frac{f(z)}{(z-a)(z-b)}dz$ around a curve in which both $a$ and $b$ lie inside.



    Should I apply Residue theorem like this:



    Residue at $a$= $frac{f(a)}{a-b}$



    Residue at $b$= $frac{f(b)}{b-a}$



    Then I use the Residue theorem 2pi*(sum of residues)



    OR



    Should I solve it by partial fractions?



    Are both valid?



    My confusion with the first method is that the numerator of $int frac{f(z)/(z-a)}{(z-b)}$, i.e. $frac{f(z)}{z-a}$ is not analytic, because $a$ also lies inside the curve. Aren't we allowed to use Cauchy's integral formula only when the numerator is analytic?






    complex-analysis residue-calculus cauchy-integral-formula analytic-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose I have to calculate $int frac{f(z)}{(z-a)(z-b)}dz$ around a curve in which both $a$ and $b$ lie inside.



      Should I apply Residue theorem like this:



      Residue at $a$= $frac{f(a)}{a-b}$



      Residue at $b$= $frac{f(b)}{b-a}$



      Then I use the Residue theorem 2pi*(sum of residues)



      OR



      Should I solve it by partial fractions?



      Are both valid?



      My confusion with the first method is that the numerator of $int frac{f(z)/(z-a)}{(z-b)}$, i.e. $frac{f(z)}{z-a}$ is not analytic, because $a$ also lies inside the curve. Aren't we allowed to use Cauchy's integral formula only when the numerator is analytic?






      complex-analysis residue-calculus cauchy-integral-formula analytic-functions







      share|cite|improve this question











      $endgroup$




      Suppose I have to calculate $int frac{f(z)}{(z-a)(z-b)}dz$ around a curve in which both $a$ and $b$ lie inside.



      Should I apply Residue theorem like this:



      Residue at $a$= $frac{f(a)}{a-b}$



      Residue at $b$= $frac{f(b)}{b-a}$



      Then I use the Residue theorem 2pi*(sum of residues)



      OR



      Should I solve it by partial fractions?



      Are both valid?



      My confusion with the first method is that the numerator of $int frac{f(z)/(z-a)}{(z-b)}$, i.e. $frac{f(z)}{z-a}$ is not analytic, because $a$ also lies inside the curve. Aren't we allowed to use Cauchy's integral formula only when the numerator is analytic?






      complex-analysis residue-calculus cauchy-integral-formula analytic-functions




      complex-analysis residue-calculus cauchy-integral-formula analytic-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 14:25









      José Carlos Santos

      162k22128232




      162k22128232










      asked Dec 20 '18 at 13:37









      Ryder RudeRyder Rude

      400111




      400111






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          They are both valid. If you apply the residue theorem, you'll get that the integral is equal to$$2pi ileft(frac{f(a)}{a-b}+frac{f(b)}{b-a}right)=2pi ifrac{f(b)-f(a)}{b-a}$$and if you apply partial fractions together with Cauchy's integral theorem, you'll getbegin{align}ointfrac{f(z)}{(z-a)(z-b)},mathrm dz&=ointfrac{f(z)}{(b-a)(z-b)}+frac{f(z)}{(a-b)(z-a)},mathrm dz\&=frac1{b-a}left(ointfrac{f(z)}{z-b},mathrm dz-ointfrac{f(z)}{z-a},mathrm dzright)\&=2pi ifrac{f(b)-f(a)}{b-a}.end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But then why are we allowed to use Cauchy's integral formula only when the numerator is analytic? Isn't integral formula just a special case of Residue theorem?
            $endgroup$
            – Ryder Rude
            Dec 20 '18 at 13:59










          • $begingroup$
            I'm a bit confused here. Cauchy's integral formula states that if $f$ is analytic, then $ointfrac{f(z)}{z-a},mathrm dz=2pi if(a)$. Isn't that what I've used (twice) in my last equality? Which hypothesis do you claim that do not hold when I applied it? And, yes, Cauchy's integral formulais a (very) special case of the residue theorem.
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 14:11











          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%2f3047546%2fwhen-to-resolve-into-partial-fractions-for-applying-cauchys-integral-formula%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          They are both valid. If you apply the residue theorem, you'll get that the integral is equal to$$2pi ileft(frac{f(a)}{a-b}+frac{f(b)}{b-a}right)=2pi ifrac{f(b)-f(a)}{b-a}$$and if you apply partial fractions together with Cauchy's integral theorem, you'll getbegin{align}ointfrac{f(z)}{(z-a)(z-b)},mathrm dz&=ointfrac{f(z)}{(b-a)(z-b)}+frac{f(z)}{(a-b)(z-a)},mathrm dz\&=frac1{b-a}left(ointfrac{f(z)}{z-b},mathrm dz-ointfrac{f(z)}{z-a},mathrm dzright)\&=2pi ifrac{f(b)-f(a)}{b-a}.end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But then why are we allowed to use Cauchy's integral formula only when the numerator is analytic? Isn't integral formula just a special case of Residue theorem?
            $endgroup$
            – Ryder Rude
            Dec 20 '18 at 13:59










          • $begingroup$
            I'm a bit confused here. Cauchy's integral formula states that if $f$ is analytic, then $ointfrac{f(z)}{z-a},mathrm dz=2pi if(a)$. Isn't that what I've used (twice) in my last equality? Which hypothesis do you claim that do not hold when I applied it? And, yes, Cauchy's integral formulais a (very) special case of the residue theorem.
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 14:11
















          1












          $begingroup$

          They are both valid. If you apply the residue theorem, you'll get that the integral is equal to$$2pi ileft(frac{f(a)}{a-b}+frac{f(b)}{b-a}right)=2pi ifrac{f(b)-f(a)}{b-a}$$and if you apply partial fractions together with Cauchy's integral theorem, you'll getbegin{align}ointfrac{f(z)}{(z-a)(z-b)},mathrm dz&=ointfrac{f(z)}{(b-a)(z-b)}+frac{f(z)}{(a-b)(z-a)},mathrm dz\&=frac1{b-a}left(ointfrac{f(z)}{z-b},mathrm dz-ointfrac{f(z)}{z-a},mathrm dzright)\&=2pi ifrac{f(b)-f(a)}{b-a}.end{align}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But then why are we allowed to use Cauchy's integral formula only when the numerator is analytic? Isn't integral formula just a special case of Residue theorem?
            $endgroup$
            – Ryder Rude
            Dec 20 '18 at 13:59










          • $begingroup$
            I'm a bit confused here. Cauchy's integral formula states that if $f$ is analytic, then $ointfrac{f(z)}{z-a},mathrm dz=2pi if(a)$. Isn't that what I've used (twice) in my last equality? Which hypothesis do you claim that do not hold when I applied it? And, yes, Cauchy's integral formulais a (very) special case of the residue theorem.
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 14:11














          1












          1








          1





          $begingroup$

          They are both valid. If you apply the residue theorem, you'll get that the integral is equal to$$2pi ileft(frac{f(a)}{a-b}+frac{f(b)}{b-a}right)=2pi ifrac{f(b)-f(a)}{b-a}$$and if you apply partial fractions together with Cauchy's integral theorem, you'll getbegin{align}ointfrac{f(z)}{(z-a)(z-b)},mathrm dz&=ointfrac{f(z)}{(b-a)(z-b)}+frac{f(z)}{(a-b)(z-a)},mathrm dz\&=frac1{b-a}left(ointfrac{f(z)}{z-b},mathrm dz-ointfrac{f(z)}{z-a},mathrm dzright)\&=2pi ifrac{f(b)-f(a)}{b-a}.end{align}






          share|cite|improve this answer











          $endgroup$



          They are both valid. If you apply the residue theorem, you'll get that the integral is equal to$$2pi ileft(frac{f(a)}{a-b}+frac{f(b)}{b-a}right)=2pi ifrac{f(b)-f(a)}{b-a}$$and if you apply partial fractions together with Cauchy's integral theorem, you'll getbegin{align}ointfrac{f(z)}{(z-a)(z-b)},mathrm dz&=ointfrac{f(z)}{(b-a)(z-b)}+frac{f(z)}{(a-b)(z-a)},mathrm dz\&=frac1{b-a}left(ointfrac{f(z)}{z-b},mathrm dz-ointfrac{f(z)}{z-a},mathrm dzright)\&=2pi ifrac{f(b)-f(a)}{b-a}.end{align}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 14:24

























          answered Dec 20 '18 at 13:46









          José Carlos SantosJosé Carlos Santos

          162k22128232




          162k22128232












          • $begingroup$
            But then why are we allowed to use Cauchy's integral formula only when the numerator is analytic? Isn't integral formula just a special case of Residue theorem?
            $endgroup$
            – Ryder Rude
            Dec 20 '18 at 13:59










          • $begingroup$
            I'm a bit confused here. Cauchy's integral formula states that if $f$ is analytic, then $ointfrac{f(z)}{z-a},mathrm dz=2pi if(a)$. Isn't that what I've used (twice) in my last equality? Which hypothesis do you claim that do not hold when I applied it? And, yes, Cauchy's integral formulais a (very) special case of the residue theorem.
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 14:11


















          • $begingroup$
            But then why are we allowed to use Cauchy's integral formula only when the numerator is analytic? Isn't integral formula just a special case of Residue theorem?
            $endgroup$
            – Ryder Rude
            Dec 20 '18 at 13:59










          • $begingroup$
            I'm a bit confused here. Cauchy's integral formula states that if $f$ is analytic, then $ointfrac{f(z)}{z-a},mathrm dz=2pi if(a)$. Isn't that what I've used (twice) in my last equality? Which hypothesis do you claim that do not hold when I applied it? And, yes, Cauchy's integral formulais a (very) special case of the residue theorem.
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 14:11
















          $begingroup$
          But then why are we allowed to use Cauchy's integral formula only when the numerator is analytic? Isn't integral formula just a special case of Residue theorem?
          $endgroup$
          – Ryder Rude
          Dec 20 '18 at 13:59




          $begingroup$
          But then why are we allowed to use Cauchy's integral formula only when the numerator is analytic? Isn't integral formula just a special case of Residue theorem?
          $endgroup$
          – Ryder Rude
          Dec 20 '18 at 13:59












          $begingroup$
          I'm a bit confused here. Cauchy's integral formula states that if $f$ is analytic, then $ointfrac{f(z)}{z-a},mathrm dz=2pi if(a)$. Isn't that what I've used (twice) in my last equality? Which hypothesis do you claim that do not hold when I applied it? And, yes, Cauchy's integral formulais a (very) special case of the residue theorem.
          $endgroup$
          – José Carlos Santos
          Dec 20 '18 at 14:11




          $begingroup$
          I'm a bit confused here. Cauchy's integral formula states that if $f$ is analytic, then $ointfrac{f(z)}{z-a},mathrm dz=2pi if(a)$. Isn't that what I've used (twice) in my last equality? Which hypothesis do you claim that do not hold when I applied it? And, yes, Cauchy's integral formulais a (very) special case of the residue theorem.
          $endgroup$
          – José Carlos Santos
          Dec 20 '18 at 14:11


















          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%2f3047546%2fwhen-to-resolve-into-partial-fractions-for-applying-cauchys-integral-formula%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