Why we can have this $int_S(0,1)exp(ipi j.x)ds=int_S(0,1)exp(ipi$?

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I can't understand why we can have this
beginalign*int_S(0,1)exp(ipi j.x)ds&=int_S(0,1)j\
&=2int_-1^1exp(ipiendalign*
with $x=(x_1,x_2),j=(j_1,j_2), |j|=sqrtj_1^2+j^2_2$, $S(0,1)$ is unit circle.



Help me, thank you.







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    up vote
    0
    down vote

    favorite












    I can't understand why we can have this
    beginalign*int_S(0,1)exp(ipi j.x)ds&=int_S(0,1)j\
    &=2int_-1^1exp(ipiendalign*
    with $x=(x_1,x_2),j=(j_1,j_2), |j|=sqrtj_1^2+j^2_2$, $S(0,1)$ is unit circle.



    Help me, thank you.







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I can't understand why we can have this
      beginalign*int_S(0,1)exp(ipi j.x)ds&=int_S(0,1)j\
      &=2int_-1^1exp(ipiendalign*
      with $x=(x_1,x_2),j=(j_1,j_2), |j|=sqrtj_1^2+j^2_2$, $S(0,1)$ is unit circle.



      Help me, thank you.







      share|cite|improve this question














      I can't understand why we can have this
      beginalign*int_S(0,1)exp(ipi j.x)ds&=int_S(0,1)j\
      &=2int_-1^1exp(ipiendalign*
      with $x=(x_1,x_2),j=(j_1,j_2), |j|=sqrtj_1^2+j^2_2$, $S(0,1)$ is unit circle.



      Help me, thank you.









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 25 at 8:36

























      asked Aug 25 at 8:17









      Chloe.Sannon

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