Prove $int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$












2












$begingroup$


Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$



(I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
$$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated






complex-analysis functional-analysis complex-numbers complex-integration







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    2












    $begingroup$


    Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$



    (I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
    $$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated






    complex-analysis functional-analysis complex-numbers complex-integration







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$



      (I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
      $$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated






      complex-analysis functional-analysis complex-numbers complex-integration







      share|cite|improve this question











      $endgroup$




      Prove: $$A=int_{0}^pi{e^{acos(x)}cos(asin(x))}dx=pi$$



      (I think) that a suggestion was made to calculate and later use it: $B=intfrac{{e^{az}}}{z}$, over the path gamma $gamma=e^{it}, tin[-pi,pi]$, i think this integral is $2ipi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral:
      $$B=int_{-pi}^pi e^{acos(x)}dx+int_{-pi}^pi e^{isin(x)}dx,$$ but do not know how to proceed. Any help appreciated






      complex-analysis functional-analysis complex-numbers complex-integration




      complex-analysis functional-analysis complex-numbers complex-integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 '18 at 13:32









      Asaf Karagila♦

      302k32427757




      302k32427757










      asked Nov 28 '18 at 13:28









      ryszard egginkryszard eggink

      308110




      308110






















          1 Answer
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          $begingroup$

          From
          $$
          2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
          $$
          we get
          $$
          int_0^{2pi}e^{ae^{it}},dt=2,pi.
          $$
          Now
          $$
          e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
          $$
          Since $2pi$ is real, we get
          $$
          int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have been there :/, this is precisely what i wrote in the last line
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 13:42












          • $begingroup$
            It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
            $endgroup$
            – Julián Aguirre
            Nov 28 '18 at 13:57










          • $begingroup$
            Ouch right on! thank you very much :)
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 14:12











          Your Answer





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          1 Answer
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          1 Answer
          1






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          active

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          active

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          2












          $begingroup$

          From
          $$
          2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
          $$
          we get
          $$
          int_0^{2pi}e^{ae^{it}},dt=2,pi.
          $$
          Now
          $$
          e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
          $$
          Since $2pi$ is real, we get
          $$
          int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have been there :/, this is precisely what i wrote in the last line
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 13:42












          • $begingroup$
            It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
            $endgroup$
            – Julián Aguirre
            Nov 28 '18 at 13:57










          • $begingroup$
            Ouch right on! thank you very much :)
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 14:12
















          2












          $begingroup$

          From
          $$
          2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
          $$
          we get
          $$
          int_0^{2pi}e^{ae^{it}},dt=2,pi.
          $$
          Now
          $$
          e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
          $$
          Since $2pi$ is real, we get
          $$
          int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have been there :/, this is precisely what i wrote in the last line
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 13:42












          • $begingroup$
            It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
            $endgroup$
            – Julián Aguirre
            Nov 28 '18 at 13:57










          • $begingroup$
            Ouch right on! thank you very much :)
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 14:12














          2












          2








          2





          $begingroup$

          From
          $$
          2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
          $$
          we get
          $$
          int_0^{2pi}e^{ae^{it}},dt=2,pi.
          $$
          Now
          $$
          e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
          $$
          Since $2pi$ is real, we get
          $$
          int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
          $$






          share|cite|improve this answer











          $endgroup$



          From
          $$
          2,pi,i=int_gammafrac{e^{iaz}}{z},dz=int_0^{2pi}frac{e^{ae^{it}}}{e^{it}},i,e^{it},dt
          $$
          we get
          $$
          int_0^{2pi}e^{ae^{it}},dt=2,pi.
          $$
          Now
          $$
          e^{ae^{it}}=e^{acos t+iasin t}=e^{acos t}bigl(cos(a,sin t)+icos(a,sin t)bigr).
          $$
          Since $2pi$ is real, we get
          $$
          int_0^{2pi}e^{acos t}cos(a,sin t),dt=2,pi.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 '18 at 13:56

























          answered Nov 28 '18 at 13:41









          Julián AguirreJulián Aguirre

          67.9k24094




          67.9k24094












          • $begingroup$
            I have been there :/, this is precisely what i wrote in the last line
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 13:42












          • $begingroup$
            It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
            $endgroup$
            – Julián Aguirre
            Nov 28 '18 at 13:57










          • $begingroup$
            Ouch right on! thank you very much :)
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 14:12


















          • $begingroup$
            I have been there :/, this is precisely what i wrote in the last line
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 13:42












          • $begingroup$
            It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
            $endgroup$
            – Julián Aguirre
            Nov 28 '18 at 13:57










          • $begingroup$
            Ouch right on! thank you very much :)
            $endgroup$
            – ryszard eggink
            Nov 28 '18 at 14:12
















          $begingroup$
          I have been there :/, this is precisely what i wrote in the last line
          $endgroup$
          – ryszard eggink
          Nov 28 '18 at 13:42






          $begingroup$
          I have been there :/, this is precisely what i wrote in the last line
          $endgroup$
          – ryszard eggink
          Nov 28 '18 at 13:42














          $begingroup$
          It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
          $endgroup$
          – Julián Aguirre
          Nov 28 '18 at 13:57




          $begingroup$
          It seems that you did not expand $e^{ae^{it}}$ correctly. Look at the edited answer.
          $endgroup$
          – Julián Aguirre
          Nov 28 '18 at 13:57












          $begingroup$
          Ouch right on! thank you very much :)
          $endgroup$
          – ryszard eggink
          Nov 28 '18 at 14:12




          $begingroup$
          Ouch right on! thank you very much :)
          $endgroup$
          – ryszard eggink
          Nov 28 '18 at 14:12


















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