Find the following integral using Cauchy's Integral Formula

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I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



$int_gamma e^zz^n dz$ where $gamma$ is the unit circle $e^itheta: -pi leq theta leq pi$ and $nin mathbfZ$.



To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



Any help will be much appreciated, thanks!!






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    up vote
    3
    down vote

    favorite












    I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



    $int_gamma e^zz^n dz$ where $gamma$ is the unit circle $e^itheta: -pi leq theta leq pi$ and $nin mathbfZ$.



    To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



    Any help will be much appreciated, thanks!!






    integration complex-analysis contour-integration complex-integration







    share|cite|improve this question









    New contributor




    Mr Bro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



      $int_gamma e^zz^n dz$ where $gamma$ is the unit circle $e^itheta: -pi leq theta leq pi$ and $nin mathbfZ$.



      To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



      Any help will be much appreciated, thanks!!






      integration complex-analysis contour-integration complex-integration







      share|cite|improve this question









      New contributor




      Mr Bro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



      $int_gamma e^zz^n dz$ where $gamma$ is the unit circle $e^itheta: -pi leq theta leq pi$ and $nin mathbfZ$.



      To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



      Any help will be much appreciated, thanks!!






      integration complex-analysis contour-integration complex-integration




      integration complex-analysis contour-integration complex-integration






      share|cite|improve this question









      New contributor




      Mr Bro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      José Carlos Santos

      124k17101187




      124k17101187






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      asked 2 hours ago









      Mr Bro

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      Mr Bro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          2 Answers
          2






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          That integral is equal to $0$ if $ngeqslant0$. In fact,beginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^n+1z,mathrm dz\&=2pi ie^00^n+1\&=0.endalignOn the other hand, if $n<0$, thenbeginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^-n,mathrm dz\&=frac2pi i(-n-1)!exp^(-n-1)(0)\&=frac2pi i(-n-1)!.endalign






          share|cite|improve this answer






















          • The question is for $n in mathbb Z$ not $n in mathbb N$.
            – Kavi Rama Murthy
            2 hours ago










          • @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
            – Mr Bro
            2 hours ago










          • @KaviRamaMurthy I've edited my answer. Thank you.
            – José Carlos Santos
            2 hours ago










          • Should there be a (−n−1)! in the denominator of the answer?
            – Mr Bro
            2 hours ago











          • @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
            – José Carlos Santos
            1 hour ago

















          up vote
          3
          down vote













          Let $f(z)=e^z$. Then $f^(k)(0)=frac k! 2pi i int_gamma frac f(z) z^k+1dz$. Use this when $n<0$ in your integral.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            That integral is equal to $0$ if $ngeqslant0$. In fact,beginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^n+1z,mathrm dz\&=2pi ie^00^n+1\&=0.endalignOn the other hand, if $n<0$, thenbeginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^-n,mathrm dz\&=frac2pi i(-n-1)!exp^(-n-1)(0)\&=frac2pi i(-n-1)!.endalign






            share|cite|improve this answer






















            • The question is for $n in mathbb Z$ not $n in mathbb N$.
              – Kavi Rama Murthy
              2 hours ago










            • @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
              – Mr Bro
              2 hours ago










            • @KaviRamaMurthy I've edited my answer. Thank you.
              – José Carlos Santos
              2 hours ago










            • Should there be a (−n−1)! in the denominator of the answer?
              – Mr Bro
              2 hours ago











            • @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
              – José Carlos Santos
              1 hour ago














            up vote
            4
            down vote



            accepted










            That integral is equal to $0$ if $ngeqslant0$. In fact,beginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^n+1z,mathrm dz\&=2pi ie^00^n+1\&=0.endalignOn the other hand, if $n<0$, thenbeginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^-n,mathrm dz\&=frac2pi i(-n-1)!exp^(-n-1)(0)\&=frac2pi i(-n-1)!.endalign






            share|cite|improve this answer






















            • The question is for $n in mathbb Z$ not $n in mathbb N$.
              – Kavi Rama Murthy
              2 hours ago










            • @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
              – Mr Bro
              2 hours ago










            • @KaviRamaMurthy I've edited my answer. Thank you.
              – José Carlos Santos
              2 hours ago










            • Should there be a (−n−1)! in the denominator of the answer?
              – Mr Bro
              2 hours ago











            • @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
              – José Carlos Santos
              1 hour ago












            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            That integral is equal to $0$ if $ngeqslant0$. In fact,beginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^n+1z,mathrm dz\&=2pi ie^00^n+1\&=0.endalignOn the other hand, if $n<0$, thenbeginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^-n,mathrm dz\&=frac2pi i(-n-1)!exp^(-n-1)(0)\&=frac2pi i(-n-1)!.endalign






            share|cite|improve this answer














            That integral is equal to $0$ if $ngeqslant0$. In fact,beginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^n+1z,mathrm dz\&=2pi ie^00^n+1\&=0.endalignOn the other hand, if $n<0$, thenbeginalignint_gamma e^zz^n,mathrm dz&=int_gammafrace^zz^-n,mathrm dz\&=frac2pi i(-n-1)!exp^(-n-1)(0)\&=frac2pi i(-n-1)!.endalign







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 21 mins ago

























            answered 2 hours ago









            José Carlos Santos

            124k17101187




            124k17101187











            • The question is for $n in mathbb Z$ not $n in mathbb N$.
              – Kavi Rama Murthy
              2 hours ago










            • @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
              – Mr Bro
              2 hours ago










            • @KaviRamaMurthy I've edited my answer. Thank you.
              – José Carlos Santos
              2 hours ago










            • Should there be a (−n−1)! in the denominator of the answer?
              – Mr Bro
              2 hours ago











            • @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
              – José Carlos Santos
              1 hour ago
















            • The question is for $n in mathbb Z$ not $n in mathbb N$.
              – Kavi Rama Murthy
              2 hours ago










            • @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
              – Mr Bro
              2 hours ago










            • @KaviRamaMurthy I've edited my answer. Thank you.
              – José Carlos Santos
              2 hours ago










            • Should there be a (−n−1)! in the denominator of the answer?
              – Mr Bro
              2 hours ago











            • @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
              – José Carlos Santos
              1 hour ago















            The question is for $n in mathbb Z$ not $n in mathbb N$.
            – Kavi Rama Murthy
            2 hours ago




            The question is for $n in mathbb Z$ not $n in mathbb N$.
            – Kavi Rama Murthy
            2 hours ago












            @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
            – Mr Bro
            2 hours ago




            @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
            – Mr Bro
            2 hours ago












            @KaviRamaMurthy I've edited my answer. Thank you.
            – José Carlos Santos
            2 hours ago




            @KaviRamaMurthy I've edited my answer. Thank you.
            – José Carlos Santos
            2 hours ago












            Should there be a (−n−1)! in the denominator of the answer?
            – Mr Bro
            2 hours ago





            Should there be a (−n−1)! in the denominator of the answer?
            – Mr Bro
            2 hours ago













            @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
            – José Carlos Santos
            1 hour ago




            @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
            – José Carlos Santos
            1 hour ago










            up vote
            3
            down vote













            Let $f(z)=e^z$. Then $f^(k)(0)=frac k! 2pi i int_gamma frac f(z) z^k+1dz$. Use this when $n<0$ in your integral.






            share|cite|improve this answer
























              up vote
              3
              down vote













              Let $f(z)=e^z$. Then $f^(k)(0)=frac k! 2pi i int_gamma frac f(z) z^k+1dz$. Use this when $n<0$ in your integral.






              share|cite|improve this answer






















                up vote
                3
                down vote










                up vote
                3
                down vote









                Let $f(z)=e^z$. Then $f^(k)(0)=frac k! 2pi i int_gamma frac f(z) z^k+1dz$. Use this when $n<0$ in your integral.






                share|cite|improve this answer












                Let $f(z)=e^z$. Then $f^(k)(0)=frac k! 2pi i int_gamma frac f(z) z^k+1dz$. Use this when $n<0$ in your integral.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Kavi Rama Murthy

                27.8k31439




                27.8k31439




















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