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







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    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






















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          $begingroup$

          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











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          • $begingroup$
            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











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          1 Answer
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          1 Answer
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          active

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          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


















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