Solving a complex integral $oint_Lfrac{e^{1/(z-a)}}zdz$ using Cauchy's formula












2












$begingroup$


I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:



$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.



So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).






integration complex-analysis complex-integration cauchy-integral-formula







share|cite|improve this question











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    2












    $begingroup$


    I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:



    $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.



    So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
    $$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
    If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).






    integration complex-analysis complex-integration cauchy-integral-formula







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:



      $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.



      So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
      $$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
      If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).






      integration complex-analysis complex-integration cauchy-integral-formula







      share|cite|improve this question











      $endgroup$




      I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:



      $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.



      So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
      $$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
      If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).






      integration complex-analysis complex-integration cauchy-integral-formula




      integration complex-analysis complex-integration cauchy-integral-formula






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 2 at 15:59









      Did

      248k23225463




      248k23225463










      asked Jan 2 at 15:56









      math101math101

      597




      597






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint



          I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.



          For a radius $r > vert a vert$, you get



          $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
            $endgroup$
            – math101
            Jan 2 at 16:37












          • $begingroup$
            I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
            $endgroup$
            – mathcounterexamples.net
            Jan 2 at 16:51












          • $begingroup$
            I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
            $endgroup$
            – math101
            Jan 2 at 17:07



















          1












          $begingroup$

          See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is the better approach.
            $endgroup$
            – Szeto
            Jan 3 at 13:32











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint



          I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.



          For a radius $r > vert a vert$, you get



          $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
            $endgroup$
            – math101
            Jan 2 at 16:37












          • $begingroup$
            I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
            $endgroup$
            – mathcounterexamples.net
            Jan 2 at 16:51












          • $begingroup$
            I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
            $endgroup$
            – math101
            Jan 2 at 17:07
















          1












          $begingroup$

          Hint



          I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.



          For a radius $r > vert a vert$, you get



          $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
            $endgroup$
            – math101
            Jan 2 at 16:37












          • $begingroup$
            I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
            $endgroup$
            – mathcounterexamples.net
            Jan 2 at 16:51












          • $begingroup$
            I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
            $endgroup$
            – math101
            Jan 2 at 17:07














          1












          1








          1





          $begingroup$

          Hint



          I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.



          For a radius $r > vert a vert$, you get



          $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$






          share|cite|improve this answer











          $endgroup$



          Hint



          I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.



          For a radius $r > vert a vert$, you get



          $$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 16:47

























          answered Jan 2 at 16:27









          mathcounterexamples.netmathcounterexamples.net

          27k22157




          27k22157












          • $begingroup$
            Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
            $endgroup$
            – math101
            Jan 2 at 16:37












          • $begingroup$
            I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
            $endgroup$
            – mathcounterexamples.net
            Jan 2 at 16:51












          • $begingroup$
            I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
            $endgroup$
            – math101
            Jan 2 at 17:07


















          • $begingroup$
            Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
            $endgroup$
            – math101
            Jan 2 at 16:37












          • $begingroup$
            I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
            $endgroup$
            – mathcounterexamples.net
            Jan 2 at 16:51












          • $begingroup$
            I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
            $endgroup$
            – math101
            Jan 2 at 17:07
















          $begingroup$
          Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
          $endgroup$
          – math101
          Jan 2 at 16:37






          $begingroup$
          Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
          $endgroup$
          – math101
          Jan 2 at 16:37














          $begingroup$
          I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
          $endgroup$
          – mathcounterexamples.net
          Jan 2 at 16:51






          $begingroup$
          I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
          $endgroup$
          – mathcounterexamples.net
          Jan 2 at 16:51














          $begingroup$
          I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
          $endgroup$
          – math101
          Jan 2 at 17:07




          $begingroup$
          I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
          $endgroup$
          – math101
          Jan 2 at 17:07











          1












          $begingroup$

          See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is the better approach.
            $endgroup$
            – Szeto
            Jan 3 at 13:32
















          1












          $begingroup$

          See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is the better approach.
            $endgroup$
            – Szeto
            Jan 3 at 13:32














          1












          1








          1





          $begingroup$

          See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.






          share|cite|improve this answer









          $endgroup$



          See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 1:34









          MaximMaxim

          5,7631220




          5,7631220












          • $begingroup$
            This is the better approach.
            $endgroup$
            – Szeto
            Jan 3 at 13:32


















          • $begingroup$
            This is the better approach.
            $endgroup$
            – Szeto
            Jan 3 at 13:32
















          $begingroup$
          This is the better approach.
          $endgroup$
          – Szeto
          Jan 3 at 13:32




          $begingroup$
          This is the better approach.
          $endgroup$
          – Szeto
          Jan 3 at 13:32


















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