Simplify integral considering only the real part












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I happened to stumble on the following simplification of an integral:



$$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$



According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?










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    1












    $begingroup$


    I happened to stumble on the following simplification of an integral:



    $$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$



    According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I happened to stumble on the following simplification of an integral:



      $$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$



      According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?










      share|cite|improve this question









      $endgroup$




      I happened to stumble on the following simplification of an integral:



      $$ frac{1}{pi} int_{0}^{infty} dx e^{-ax} cdot cos(kx) = frac{1}{pi} Re left[ int_{0}^{infty} dx e^{x (ik - a)} right] $$



      According to my reasoning: $ cos(kx) = frac{1}{2} left( e^{ikx} + e^{-ikx} right) $, so the term on the exponent on the right side comes somehow from the first term of the complex form of $cos(kx)$, but the second term gets cancelled, why?







      integration complex-analysis analysis complex-integration






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      asked Jan 12 at 8:45









      LeroyLeroy

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      1267






















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          $begingroup$

          Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
          $$
          Re(e^{ikx}) = cos(kx).
          $$






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            0












            $begingroup$

            Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
            $$
            Re(e^{ikx}) = cos(kx).
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
              $$
              Re(e^{ikx}) = cos(kx).
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
                $$
                Re(e^{ikx}) = cos(kx).
                $$






                share|cite|improve this answer









                $endgroup$



                Consider first $e^{ikx} = cos(kx) + i sin(kx)$ then you can easily deduce that
                $$
                Re(e^{ikx}) = cos(kx).
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 12 at 8:59









                WarreGWarreG

                20819




                20819






























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