Integration of two exponential multiplied by each other












2












$begingroup$


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







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  • $begingroup$
    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
    1 hour ago
















2












$begingroup$


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
    1 hour ago














2












2








2





$begingroup$


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






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share|cite|improve this question













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share|cite|improve this question








edited 2 hours ago









Thomas Shelby

3,6342525




3,6342525










asked 3 hours ago









articatarticat

163




163












  • $begingroup$
    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
    1 hour 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
    1 hour 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
1 hour 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
1 hour ago










2 Answers
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$begingroup$

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






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    2












    $begingroup$

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






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      2 Answers
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      2 Answers
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      $begingroup$

      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











      $endgroup$


















        4












        $begingroup$

        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











        $endgroup$
















          4












          4








          4





          $begingroup$

          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 3 hours ago

























          answered 3 hours ago









          Thomas ShelbyThomas Shelby

          3,6342525




          3,6342525























              2












              $begingroup$

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






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

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






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 3 hours ago









                  Graham KempGraham Kemp

                  86.1k43478




                  86.1k43478






























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