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







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












$begingroup$


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 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
2






active

oldest

votes


















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











$endgroup$





















    2












    $begingroup$

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






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3126893%2fintegration-of-two-exponential-multiplied-by-each-other%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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












        $begingroup$

        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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3126893%2fintegration-of-two-exponential-multiplied-by-each-other%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Knooppunt Holsloot

                      Image
                      Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
                      Read more

                      Altaar (religie)

                      Image
                      Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
                      Read more

                      Gregoriusmis

                      Image
                      Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
                      Read more