Compute the integral $ int_{-pi}^{pi}e^{e^{ix}}e^{-ikx}text{d}x $












1












$begingroup$


I was told to compute the following integral
$$
int_{-pi}^{pi}e^{e^{ix}}e^{-ikx}text{d}x
$$
where $kinmathbb{Z}$ but I don't know how to start. Should I expand the exponentials in terms of trigonometric functions or what is the trick here? Can I use theory about Fourier series to simplify this? Any help is welcome.






integration definite-integrals exponential-function







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$endgroup$












  • $begingroup$
    Were you told whether or not you'd get an exact answer?
    $endgroup$
    – Cardioid_Ass_22
    Jan 28 at 18:35










  • $begingroup$
    Do you know enough complex analysis to know the Cauchy Integral Formula?
    $endgroup$
    – JimmyK4542
    Jan 28 at 18:38










  • $begingroup$
    I was told to compute the precise value of that integral for all $kinmathbb{Z}$ so I guess there is an exact answer.
    $endgroup$
    – peastick
    Jan 28 at 18:39












  • $begingroup$
    @JimmyK4542 I know that formula, but this is an exercise in Fourier analysis course.
    $endgroup$
    – peastick
    Jan 28 at 18:41
















1












$begingroup$


I was told to compute the following integral
$$
int_{-pi}^{pi}e^{e^{ix}}e^{-ikx}text{d}x
$$
where $kinmathbb{Z}$ but I don't know how to start. Should I expand the exponentials in terms of trigonometric functions or what is the trick here? Can I use theory about Fourier series to simplify this? Any help is welcome.






integration definite-integrals exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Were you told whether or not you'd get an exact answer?
    $endgroup$
    – Cardioid_Ass_22
    Jan 28 at 18:35










  • $begingroup$
    Do you know enough complex analysis to know the Cauchy Integral Formula?
    $endgroup$
    – JimmyK4542
    Jan 28 at 18:38










  • $begingroup$
    I was told to compute the precise value of that integral for all $kinmathbb{Z}$ so I guess there is an exact answer.
    $endgroup$
    – peastick
    Jan 28 at 18:39












  • $begingroup$
    @JimmyK4542 I know that formula, but this is an exercise in Fourier analysis course.
    $endgroup$
    – peastick
    Jan 28 at 18:41














1












1








1





$begingroup$


I was told to compute the following integral
$$
int_{-pi}^{pi}e^{e^{ix}}e^{-ikx}text{d}x
$$
where $kinmathbb{Z}$ but I don't know how to start. Should I expand the exponentials in terms of trigonometric functions or what is the trick here? Can I use theory about Fourier series to simplify this? Any help is welcome.






integration definite-integrals exponential-function







share|cite|improve this question











$endgroup$




I was told to compute the following integral
$$
int_{-pi}^{pi}e^{e^{ix}}e^{-ikx}text{d}x
$$
where $kinmathbb{Z}$ but I don't know how to start. Should I expand the exponentials in terms of trigonometric functions or what is the trick here? Can I use theory about Fourier series to simplify this? Any help is welcome.






integration definite-integrals exponential-function




integration definite-integrals exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 18:40







peastick

















asked Jan 28 at 18:32









peastickpeastick

10718




10718












  • $begingroup$
    Were you told whether or not you'd get an exact answer?
    $endgroup$
    – Cardioid_Ass_22
    Jan 28 at 18:35










  • $begingroup$
    Do you know enough complex analysis to know the Cauchy Integral Formula?
    $endgroup$
    – JimmyK4542
    Jan 28 at 18:38










  • $begingroup$
    I was told to compute the precise value of that integral for all $kinmathbb{Z}$ so I guess there is an exact answer.
    $endgroup$
    – peastick
    Jan 28 at 18:39












  • $begingroup$
    @JimmyK4542 I know that formula, but this is an exercise in Fourier analysis course.
    $endgroup$
    – peastick
    Jan 28 at 18:41


















  • $begingroup$
    Were you told whether or not you'd get an exact answer?
    $endgroup$
    – Cardioid_Ass_22
    Jan 28 at 18:35










  • $begingroup$
    Do you know enough complex analysis to know the Cauchy Integral Formula?
    $endgroup$
    – JimmyK4542
    Jan 28 at 18:38










  • $begingroup$
    I was told to compute the precise value of that integral for all $kinmathbb{Z}$ so I guess there is an exact answer.
    $endgroup$
    – peastick
    Jan 28 at 18:39












  • $begingroup$
    @JimmyK4542 I know that formula, but this is an exercise in Fourier analysis course.
    $endgroup$
    – peastick
    Jan 28 at 18:41
















$begingroup$
Were you told whether or not you'd get an exact answer?
$endgroup$
– Cardioid_Ass_22
Jan 28 at 18:35




$begingroup$
Were you told whether or not you'd get an exact answer?
$endgroup$
– Cardioid_Ass_22
Jan 28 at 18:35












$begingroup$
Do you know enough complex analysis to know the Cauchy Integral Formula?
$endgroup$
– JimmyK4542
Jan 28 at 18:38




$begingroup$
Do you know enough complex analysis to know the Cauchy Integral Formula?
$endgroup$
– JimmyK4542
Jan 28 at 18:38












$begingroup$
I was told to compute the precise value of that integral for all $kinmathbb{Z}$ so I guess there is an exact answer.
$endgroup$
– peastick
Jan 28 at 18:39






$begingroup$
I was told to compute the precise value of that integral for all $kinmathbb{Z}$ so I guess there is an exact answer.
$endgroup$
– peastick
Jan 28 at 18:39














$begingroup$
@JimmyK4542 I know that formula, but this is an exercise in Fourier analysis course.
$endgroup$
– peastick
Jan 28 at 18:41




$begingroup$
@JimmyK4542 I know that formula, but this is an exercise in Fourier analysis course.
$endgroup$
– peastick
Jan 28 at 18:41










3 Answers
3






active

oldest

votes


















3












$begingroup$

Write
$$e^{e^{ix}}=sum_{n=0}^infty frac{e^{nix}}{n!}$$
and use the fact that
$$int_{-pi}^pi e^{nix} , dx=
begin{cases}2pi & n = 0\
0 & text{otherwise}
end{cases}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Am I allowed to change the order of summation and integration? If that is true, I get $frac{2pi}{k!}$ as an answer.
    $endgroup$
    – peastick
    Jan 28 at 18:52








  • 1




    $begingroup$
    I don't know what theorems you know at the moment, but hopefully you have some way to justify interchanges, as it would be difficult to do Fourier analysis without one! I personally would probably use the bounded convergence theorem.
    $endgroup$
    – Micah
    Jan 28 at 19:07










  • $begingroup$
    Also, beware of negative $k$ in your answer...
    $endgroup$
    – Micah
    Jan 28 at 19:07



















2












$begingroup$

Hint: Let $z=e^{ix}$. The integral becomes a contour integral along the unit circle, so Cauchy integral formula applies.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Put



    $$z:=e^{ix};,;;-pile xle piimplies dz=iz,dx,,,,|z|=1implies dx=-frac izdzimplies$$



    $$int_{-pi}^pi e^{e^{ix}}e^{ikx},dx=oint_C e^zz^kleft(-frac iz,dzright)=frac1ioint_Ce^zz^{k-1}dz=
    begin{cases}0,&kge1\{}\
    cfrac{2pi}{(-k)!},&k<1end{cases}$$



    With $;C=;$ the unit circle. Use Cauchy's Integral Formula






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Write
      $$e^{e^{ix}}=sum_{n=0}^infty frac{e^{nix}}{n!}$$
      and use the fact that
      $$int_{-pi}^pi e^{nix} , dx=
      begin{cases}2pi & n = 0\
      0 & text{otherwise}
      end{cases}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Am I allowed to change the order of summation and integration? If that is true, I get $frac{2pi}{k!}$ as an answer.
        $endgroup$
        – peastick
        Jan 28 at 18:52








      • 1




        $begingroup$
        I don't know what theorems you know at the moment, but hopefully you have some way to justify interchanges, as it would be difficult to do Fourier analysis without one! I personally would probably use the bounded convergence theorem.
        $endgroup$
        – Micah
        Jan 28 at 19:07










      • $begingroup$
        Also, beware of negative $k$ in your answer...
        $endgroup$
        – Micah
        Jan 28 at 19:07
















      3












      $begingroup$

      Write
      $$e^{e^{ix}}=sum_{n=0}^infty frac{e^{nix}}{n!}$$
      and use the fact that
      $$int_{-pi}^pi e^{nix} , dx=
      begin{cases}2pi & n = 0\
      0 & text{otherwise}
      end{cases}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Am I allowed to change the order of summation and integration? If that is true, I get $frac{2pi}{k!}$ as an answer.
        $endgroup$
        – peastick
        Jan 28 at 18:52








      • 1




        $begingroup$
        I don't know what theorems you know at the moment, but hopefully you have some way to justify interchanges, as it would be difficult to do Fourier analysis without one! I personally would probably use the bounded convergence theorem.
        $endgroup$
        – Micah
        Jan 28 at 19:07










      • $begingroup$
        Also, beware of negative $k$ in your answer...
        $endgroup$
        – Micah
        Jan 28 at 19:07














      3












      3








      3





      $begingroup$

      Write
      $$e^{e^{ix}}=sum_{n=0}^infty frac{e^{nix}}{n!}$$
      and use the fact that
      $$int_{-pi}^pi e^{nix} , dx=
      begin{cases}2pi & n = 0\
      0 & text{otherwise}
      end{cases}$$






      share|cite|improve this answer









      $endgroup$



      Write
      $$e^{e^{ix}}=sum_{n=0}^infty frac{e^{nix}}{n!}$$
      and use the fact that
      $$int_{-pi}^pi e^{nix} , dx=
      begin{cases}2pi & n = 0\
      0 & text{otherwise}
      end{cases}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 28 at 18:48









      MicahMicah

      30.2k1364106




      30.2k1364106












      • $begingroup$
        Am I allowed to change the order of summation and integration? If that is true, I get $frac{2pi}{k!}$ as an answer.
        $endgroup$
        – peastick
        Jan 28 at 18:52








      • 1




        $begingroup$
        I don't know what theorems you know at the moment, but hopefully you have some way to justify interchanges, as it would be difficult to do Fourier analysis without one! I personally would probably use the bounded convergence theorem.
        $endgroup$
        – Micah
        Jan 28 at 19:07










      • $begingroup$
        Also, beware of negative $k$ in your answer...
        $endgroup$
        – Micah
        Jan 28 at 19:07


















      • $begingroup$
        Am I allowed to change the order of summation and integration? If that is true, I get $frac{2pi}{k!}$ as an answer.
        $endgroup$
        – peastick
        Jan 28 at 18:52








      • 1




        $begingroup$
        I don't know what theorems you know at the moment, but hopefully you have some way to justify interchanges, as it would be difficult to do Fourier analysis without one! I personally would probably use the bounded convergence theorem.
        $endgroup$
        – Micah
        Jan 28 at 19:07










      • $begingroup$
        Also, beware of negative $k$ in your answer...
        $endgroup$
        – Micah
        Jan 28 at 19:07
















      $begingroup$
      Am I allowed to change the order of summation and integration? If that is true, I get $frac{2pi}{k!}$ as an answer.
      $endgroup$
      – peastick
      Jan 28 at 18:52






      $begingroup$
      Am I allowed to change the order of summation and integration? If that is true, I get $frac{2pi}{k!}$ as an answer.
      $endgroup$
      – peastick
      Jan 28 at 18:52






      1




      1




      $begingroup$
      I don't know what theorems you know at the moment, but hopefully you have some way to justify interchanges, as it would be difficult to do Fourier analysis without one! I personally would probably use the bounded convergence theorem.
      $endgroup$
      – Micah
      Jan 28 at 19:07




      $begingroup$
      I don't know what theorems you know at the moment, but hopefully you have some way to justify interchanges, as it would be difficult to do Fourier analysis without one! I personally would probably use the bounded convergence theorem.
      $endgroup$
      – Micah
      Jan 28 at 19:07












      $begingroup$
      Also, beware of negative $k$ in your answer...
      $endgroup$
      – Micah
      Jan 28 at 19:07




      $begingroup$
      Also, beware of negative $k$ in your answer...
      $endgroup$
      – Micah
      Jan 28 at 19:07











      2












      $begingroup$

      Hint: Let $z=e^{ix}$. The integral becomes a contour integral along the unit circle, so Cauchy integral formula applies.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint: Let $z=e^{ix}$. The integral becomes a contour integral along the unit circle, so Cauchy integral formula applies.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint: Let $z=e^{ix}$. The integral becomes a contour integral along the unit circle, so Cauchy integral formula applies.






          share|cite|improve this answer









          $endgroup$



          Hint: Let $z=e^{ix}$. The integral becomes a contour integral along the unit circle, so Cauchy integral formula applies.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 28 at 18:45









          Eclipse SunEclipse Sun

          7,9851438




          7,9851438























              1












              $begingroup$

              Put



              $$z:=e^{ix};,;;-pile xle piimplies dz=iz,dx,,,,|z|=1implies dx=-frac izdzimplies$$



              $$int_{-pi}^pi e^{e^{ix}}e^{ikx},dx=oint_C e^zz^kleft(-frac iz,dzright)=frac1ioint_Ce^zz^{k-1}dz=
              begin{cases}0,&kge1\{}\
              cfrac{2pi}{(-k)!},&k<1end{cases}$$



              With $;C=;$ the unit circle. Use Cauchy's Integral Formula






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Put



                $$z:=e^{ix};,;;-pile xle piimplies dz=iz,dx,,,,|z|=1implies dx=-frac izdzimplies$$



                $$int_{-pi}^pi e^{e^{ix}}e^{ikx},dx=oint_C e^zz^kleft(-frac iz,dzright)=frac1ioint_Ce^zz^{k-1}dz=
                begin{cases}0,&kge1\{}\
                cfrac{2pi}{(-k)!},&k<1end{cases}$$



                With $;C=;$ the unit circle. Use Cauchy's Integral Formula






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Put



                  $$z:=e^{ix};,;;-pile xle piimplies dz=iz,dx,,,,|z|=1implies dx=-frac izdzimplies$$



                  $$int_{-pi}^pi e^{e^{ix}}e^{ikx},dx=oint_C e^zz^kleft(-frac iz,dzright)=frac1ioint_Ce^zz^{k-1}dz=
                  begin{cases}0,&kge1\{}\
                  cfrac{2pi}{(-k)!},&k<1end{cases}$$



                  With $;C=;$ the unit circle. Use Cauchy's Integral Formula






                  share|cite|improve this answer









                  $endgroup$



                  Put



                  $$z:=e^{ix};,;;-pile xle piimplies dz=iz,dx,,,,|z|=1implies dx=-frac izdzimplies$$



                  $$int_{-pi}^pi e^{e^{ix}}e^{ikx},dx=oint_C e^zz^kleft(-frac iz,dzright)=frac1ioint_Ce^zz^{k-1}dz=
                  begin{cases}0,&kge1\{}\
                  cfrac{2pi}{(-k)!},&k<1end{cases}$$



                  With $;C=;$ the unit circle. Use Cauchy's Integral Formula







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 28 at 18:54









                  DonAntonioDonAntonio

                  180k1494233




                  180k1494233






























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