Don't understand an integral with complex numbers.












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I am studying the fourier transformation and I don't understand this integral (it is set to find the Fourier coefficients of the function $f(x)=x$: integralSpecifically, I know it is integrating by parts, but I don't understand the last equality. I'd appreciate if someone could elaborate. Thank you






integration fourier-series







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  • 1




    $begingroup$
    I recommend taking the expression before the last $=$ and trying to evaluate it yourself.
    $endgroup$
    – Wojowu
    Jan 27 at 13:58
















0












$begingroup$


I am studying the fourier transformation and I don't understand this integral (it is set to find the Fourier coefficients of the function $f(x)=x$: integralSpecifically, I know it is integrating by parts, but I don't understand the last equality. I'd appreciate if someone could elaborate. Thank you






integration fourier-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I recommend taking the expression before the last $=$ and trying to evaluate it yourself.
    $endgroup$
    – Wojowu
    Jan 27 at 13:58














0












0








0





$begingroup$


I am studying the fourier transformation and I don't understand this integral (it is set to find the Fourier coefficients of the function $f(x)=x$: integralSpecifically, I know it is integrating by parts, but I don't understand the last equality. I'd appreciate if someone could elaborate. Thank you






integration fourier-series







share|cite|improve this question









$endgroup$




I am studying the fourier transformation and I don't understand this integral (it is set to find the Fourier coefficients of the function $f(x)=x$: integralSpecifically, I know it is integrating by parts, but I don't understand the last equality. I'd appreciate if someone could elaborate. Thank you






integration fourier-series




integration fourier-series






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asked Jan 27 at 13:57









codingnightcodingnight

877




877








  • 1




    $begingroup$
    I recommend taking the expression before the last $=$ and trying to evaluate it yourself.
    $endgroup$
    – Wojowu
    Jan 27 at 13:58














  • 1




    $begingroup$
    I recommend taking the expression before the last $=$ and trying to evaluate it yourself.
    $endgroup$
    – Wojowu
    Jan 27 at 13:58








1




1




$begingroup$
I recommend taking the expression before the last $=$ and trying to evaluate it yourself.
$endgroup$
– Wojowu
Jan 27 at 13:58




$begingroup$
I recommend taking the expression before the last $=$ and trying to evaluate it yourself.
$endgroup$
– Wojowu
Jan 27 at 13:58










2 Answers
2






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1












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First of all, note thatbegin{align}int_{-frac12}^frac12e^{-2pi inx},mathrm dx&=left[frac{e^{-2pi inx}}{2pi in}right]_{x=-frac12}^{x=frac12}=0,end{align}since the exponential function is periodic with period $2pi i$.



So, the expression after the second $=$ sign is just$$left[frac{-1}{2pi in}e^{-2pi inx}right]_{x=-frac12}^{x=frac12},$$which is precisely $dfrac{(-1)^{n+1}}{2pi in}$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Hint:



    Expanding out the RHS gives us:



    $$frac{1}{2nipi}bigg(frac 12(e^{npi i}+e^{-npi i})bigg)+frac{1}{2nipi}bigg(frac{1}{2nipi}(e^{-npi i}-e^{npi i})bigg)$$



    Can you see this is:



    $$frac{cos (npi)}{2nipi}+frac{sin(-npi)}{2n^2pi^2i}$$






    share|cite|improve this answer









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      Your Answer





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      2 Answers
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      2 Answers
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      1












      $begingroup$

      First of all, note thatbegin{align}int_{-frac12}^frac12e^{-2pi inx},mathrm dx&=left[frac{e^{-2pi inx}}{2pi in}right]_{x=-frac12}^{x=frac12}=0,end{align}since the exponential function is periodic with period $2pi i$.



      So, the expression after the second $=$ sign is just$$left[frac{-1}{2pi in}e^{-2pi inx}right]_{x=-frac12}^{x=frac12},$$which is precisely $dfrac{(-1)^{n+1}}{2pi in}$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        First of all, note thatbegin{align}int_{-frac12}^frac12e^{-2pi inx},mathrm dx&=left[frac{e^{-2pi inx}}{2pi in}right]_{x=-frac12}^{x=frac12}=0,end{align}since the exponential function is periodic with period $2pi i$.



        So, the expression after the second $=$ sign is just$$left[frac{-1}{2pi in}e^{-2pi inx}right]_{x=-frac12}^{x=frac12},$$which is precisely $dfrac{(-1)^{n+1}}{2pi in}$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          First of all, note thatbegin{align}int_{-frac12}^frac12e^{-2pi inx},mathrm dx&=left[frac{e^{-2pi inx}}{2pi in}right]_{x=-frac12}^{x=frac12}=0,end{align}since the exponential function is periodic with period $2pi i$.



          So, the expression after the second $=$ sign is just$$left[frac{-1}{2pi in}e^{-2pi inx}right]_{x=-frac12}^{x=frac12},$$which is precisely $dfrac{(-1)^{n+1}}{2pi in}$.






          share|cite|improve this answer











          $endgroup$



          First of all, note thatbegin{align}int_{-frac12}^frac12e^{-2pi inx},mathrm dx&=left[frac{e^{-2pi inx}}{2pi in}right]_{x=-frac12}^{x=frac12}=0,end{align}since the exponential function is periodic with period $2pi i$.



          So, the expression after the second $=$ sign is just$$left[frac{-1}{2pi in}e^{-2pi inx}right]_{x=-frac12}^{x=frac12},$$which is precisely $dfrac{(-1)^{n+1}}{2pi in}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 27 at 14:24

























          answered Jan 27 at 14:08









          José Carlos SantosJosé Carlos Santos

          170k23132238




          170k23132238























              0












              $begingroup$

              Hint:



              Expanding out the RHS gives us:



              $$frac{1}{2nipi}bigg(frac 12(e^{npi i}+e^{-npi i})bigg)+frac{1}{2nipi}bigg(frac{1}{2nipi}(e^{-npi i}-e^{npi i})bigg)$$



              Can you see this is:



              $$frac{cos (npi)}{2nipi}+frac{sin(-npi)}{2n^2pi^2i}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint:



                Expanding out the RHS gives us:



                $$frac{1}{2nipi}bigg(frac 12(e^{npi i}+e^{-npi i})bigg)+frac{1}{2nipi}bigg(frac{1}{2nipi}(e^{-npi i}-e^{npi i})bigg)$$



                Can you see this is:



                $$frac{cos (npi)}{2nipi}+frac{sin(-npi)}{2n^2pi^2i}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:



                  Expanding out the RHS gives us:



                  $$frac{1}{2nipi}bigg(frac 12(e^{npi i}+e^{-npi i})bigg)+frac{1}{2nipi}bigg(frac{1}{2nipi}(e^{-npi i}-e^{npi i})bigg)$$



                  Can you see this is:



                  $$frac{cos (npi)}{2nipi}+frac{sin(-npi)}{2n^2pi^2i}$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  Expanding out the RHS gives us:



                  $$frac{1}{2nipi}bigg(frac 12(e^{npi i}+e^{-npi i})bigg)+frac{1}{2nipi}bigg(frac{1}{2nipi}(e^{-npi i}-e^{npi i})bigg)$$



                  Can you see this is:



                  $$frac{cos (npi)}{2nipi}+frac{sin(-npi)}{2n^2pi^2i}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 27 at 14:18









                  Rhys HughesRhys Hughes

                  7,0801630




                  7,0801630






























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