A problem when integrate Cos[n*x]*Cos[k*x]












3












$begingroup$


When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code 



sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi}, 

  Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


gives the result (k Sin[π k + π n] - n Sin[π k + π n] +
k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2).
enter image description here



And then use the Simplify function, 



Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?






calculus-and-analysis







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    3












    $begingroup$


    When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code 



    sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi}, 
    
  Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


    gives the result (k Sin[π k + π n] - n Sin[π k + π n] +
    k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2)    .
    enter image description here



    And then use the Simplify function, 



    Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


    gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?






    calculus-and-analysis







    share|improve this question









    New contributor




    shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3


      1



      $begingroup$


      When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code 



      sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi}, 
      
  Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


      gives the result (k Sin[π k + π n] - n Sin[π k + π n] +
      k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2)      .
      enter image description here



      And then use the Simplify function, 



      Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


      gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?






      calculus-and-analysis







      share|improve this question









      New contributor




      shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code 



      sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi}, 
      
  Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


      gives the result (k Sin[π k + π n] - n Sin[π k + π n] +
      k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2)      .
      enter image description here



      And then use the Simplify function, 



      Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]


      gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?






      calculus-and-analysis




      calculus-and-analysis






      share|improve this question









      New contributor




      shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 2 hours ago









      Mr.Wizard♦

      231k294751042




      231k294751042






      New contributor




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      asked 4 hours ago









      shelure21shelure21

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      shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          1 Answer
          1






          active

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          4












          $begingroup$

          This is well know issue. One way to handle it is 



          Simplify[ sol, 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
          

          
(* 0 *)


          And



          Simplify[ Limit[sol, k -> n], 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
          

          
(* Pi *)


          See



          should-integrate-detect-orthogonality-of-functions-in-the-integrand



          And



          What assumptions to use to check for orthogonality



          And



          should-integrate-have-given-zero-for-this-integral



          And



          proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints



          And



          usage-of-assuming-for-integration






          share|improve this answer









          $endgroup$













          • $begingroup$
            You can shorten the Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]
            $endgroup$
            – Bob Hanlon
            43 mins ago










          • $begingroup$
            @BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
            $endgroup$
            – Nasser
            28 mins ago











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          1 Answer
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          4












          $begingroup$

          This is well know issue. One way to handle it is 



          Simplify[ sol, 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
          

          
(* 0 *)


          And



          Simplify[ Limit[sol, k -> n], 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
          

          
(* Pi *)


          See



          should-integrate-detect-orthogonality-of-functions-in-the-integrand



          And



          What assumptions to use to check for orthogonality



          And



          should-integrate-have-given-zero-for-this-integral



          And



          proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints



          And



          usage-of-assuming-for-integration






          share|improve this answer









          $endgroup$













          • $begingroup$
            You can shorten the Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]
            $endgroup$
            – Bob Hanlon
            43 mins ago










          • $begingroup$
            @BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
            $endgroup$
            – Nasser
            28 mins ago
















          4












          $begingroup$

          This is well know issue. One way to handle it is 



          Simplify[ sol, 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
          

          
(* 0 *)


          And



          Simplify[ Limit[sol, k -> n], 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
          

          
(* Pi *)


          See



          should-integrate-detect-orthogonality-of-functions-in-the-integrand



          And



          What assumptions to use to check for orthogonality



          And



          should-integrate-have-given-zero-for-this-integral



          And



          proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints



          And



          usage-of-assuming-for-integration






          share|improve this answer









          $endgroup$













          • $begingroup$
            You can shorten the Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]
            $endgroup$
            – Bob Hanlon
            43 mins ago










          • $begingroup$
            @BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
            $endgroup$
            – Nasser
            28 mins ago














          4












          4








          4





          $begingroup$

          This is well know issue. One way to handle it is 



          Simplify[ sol, 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
          

          
(* 0 *)


          And



          Simplify[ Limit[sol, k -> n], 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
          

          
(* Pi *)


          See



          should-integrate-detect-orthogonality-of-functions-in-the-integrand



          And



          What assumptions to use to check for orthogonality



          And



          should-integrate-have-given-zero-for-this-integral



          And



          proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints



          And



          usage-of-assuming-for-integration






          share|improve this answer









          $endgroup$



          This is well know issue. One way to handle it is 



          Simplify[ sol, 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
          

          
(* 0 *)


          And



          Simplify[ Limit[sol, k -> n], 
          
 Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
          

          
(* Pi *)


          See



          should-integrate-detect-orthogonality-of-functions-in-the-integrand



          And



          What assumptions to use to check for orthogonality



          And



          should-integrate-have-given-zero-for-this-integral



          And



          proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints



          And



          usage-of-assuming-for-integration







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          NasserNasser

          57.7k488205




          57.7k488205












          • $begingroup$
            You can shorten the Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]
            $endgroup$
            – Bob Hanlon
            43 mins ago










          • $begingroup$
            @BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
            $endgroup$
            – Nasser
            28 mins ago


















          • $begingroup$
            You can shorten the Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]
            $endgroup$
            – Bob Hanlon
            43 mins ago










          • $begingroup$
            @BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
            $endgroup$
            – Nasser
            28 mins ago
















          $begingroup$
          You can shorten the Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]
          $endgroup$
          – Bob Hanlon
          43 mins ago




          $begingroup$
          You can shorten the Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]
          $endgroup$
          – Bob Hanlon
          43 mins ago












          $begingroup$
          @BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
          $endgroup$
          – Nasser
          28 mins ago




          $begingroup$
          @BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
          $endgroup$
          – Nasser
          28 mins ago










          shelure21 is a new contributor. Be nice, and check out our Code of Conduct.










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