Integration of two exponential multiplied by each other












3












$begingroup$


I am having confusion on how to go about integrating this integral:



$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$



I attempted by using integration by parts but that didn't work.






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
    $endgroup$
    – Gus
    3 hours ago
















3












$begingroup$


I am having confusion on how to go about integrating this integral:



$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$



I attempted by using integration by parts but that didn't work.






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
    $endgroup$
    – Gus
    3 hours ago














3












3








3





$begingroup$


I am having confusion on how to go about integrating this integral:



$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$



I attempted by using integration by parts but that didn't work.






integration







share|cite|improve this question











$endgroup$




I am having confusion on how to go about integrating this integral:



$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$



I attempted by using integration by parts but that didn't work.






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Thomas Shelby

3,6642525




3,6642525










asked 5 hours ago









articatarticat

213




213












  • $begingroup$
    This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
    $endgroup$
    – Gus
    3 hours ago


















  • $begingroup$
    This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
    $endgroup$
    – Gus
    3 hours ago
















$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
$endgroup$
– Gus
3 hours ago




$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
$endgroup$
– Gus
3 hours ago










2 Answers
2






active

oldest

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6












$begingroup$

Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      exponentiation rules - yes it does
      $endgroup$
      – qwr
      6 mins ago











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    6












    $begingroup$

    Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






        share|cite|improve this answer











        $endgroup$



        Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 5 hours ago

























        answered 5 hours ago









        Thomas ShelbyThomas Shelby

        3,6642525




        3,6642525























            3












            $begingroup$

            Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              exponentiation rules - yes it does
              $endgroup$
              – qwr
              6 mins ago
















            3












            $begingroup$

            Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              exponentiation rules - yes it does
              $endgroup$
              – qwr
              6 mins ago














            3












            3








            3





            $begingroup$

            Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






            share|cite|improve this answer









            $endgroup$



            Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 5 hours ago









            Graham KempGraham Kemp

            86.1k43478




            86.1k43478












            • $begingroup$
              exponentiation rules - yes it does
              $endgroup$
              – qwr
              6 mins ago


















            • $begingroup$
              exponentiation rules - yes it does
              $endgroup$
              – qwr
              6 mins ago
















            $begingroup$
            exponentiation rules - yes it does
            $endgroup$
            – qwr
            6 mins ago




            $begingroup$
            exponentiation rules - yes it does
            $endgroup$
            – qwr
            6 mins ago


















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