Fourier Transforms of functions on a finite interval











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I would like to find the fourier transform of $sin(ax)$ which is non zero only on $-cle x le c$.



Calculating, I get an answer with sines and cosines but I'm not sure if this is correct... Should I be using the delta function in my answer? Or is that if the interval is infinite?






fourier-transform







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    up vote
    -2
    down vote

    favorite












    I would like to find the fourier transform of $sin(ax)$ which is non zero only on $-cle x le c$.



    Calculating, I get an answer with sines and cosines but I'm not sure if this is correct... Should I be using the delta function in my answer? Or is that if the interval is infinite?






    fourier-transform







    share|cite|improve this question


























      up vote
      -2
      down vote

      favorite









      up vote
      -2
      down vote

      favorite











      I would like to find the fourier transform of $sin(ax)$ which is non zero only on $-cle x le c$.



      Calculating, I get an answer with sines and cosines but I'm not sure if this is correct... Should I be using the delta function in my answer? Or is that if the interval is infinite?






      fourier-transform







      share|cite|improve this question















      I would like to find the fourier transform of $sin(ax)$ which is non zero only on $-cle x le c$.



      Calculating, I get an answer with sines and cosines but I'm not sure if this is correct... Should I be using the delta function in my answer? Or is that if the interval is infinite?






      fourier-transform




      fourier-transform






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      share|cite|improve this question













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      edited 1 hour ago









      Tianlalu

      2,170631




      2,170631










      asked 2 hours ago









      MathematicianP

      3115




      3115






















          1 Answer
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          You can represent your function as
          $$
          f(x)=[theta(x+c)-theta(x-c)]sin(ax)
          $$
          being $theta(x)$ the Heaviside step function. Then, the Fourier integral becomes rather easy as
          $$
          {cal F}(f)(k)=int_{-infty}^infty dx e^{ikx}f(x)=int_{-c}^c dx e^{ikx}sin(ax).
          $$
          The latter integral is very well-known and rather easy to evaluate.






          share|cite|improve this answer





















          • Thank you for your reply! I get an integral involving sin(ac) and cos(kc),does this sound correct?
            – MathematicianP
            1 hour ago










          • Yes, it is correct.
            – Jon
            1 hour ago










          • Perfect! Thank you so much!
            – MathematicianP
            1 hour ago










          • If you like the answer, please accept it.
            – Jon
            1 hour ago











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote



          accepted










          You can represent your function as
          $$
          f(x)=[theta(x+c)-theta(x-c)]sin(ax)
          $$
          being $theta(x)$ the Heaviside step function. Then, the Fourier integral becomes rather easy as
          $$
          {cal F}(f)(k)=int_{-infty}^infty dx e^{ikx}f(x)=int_{-c}^c dx e^{ikx}sin(ax).
          $$
          The latter integral is very well-known and rather easy to evaluate.






          share|cite|improve this answer





















          • Thank you for your reply! I get an integral involving sin(ac) and cos(kc),does this sound correct?
            – MathematicianP
            1 hour ago










          • Yes, it is correct.
            – Jon
            1 hour ago










          • Perfect! Thank you so much!
            – MathematicianP
            1 hour ago










          • If you like the answer, please accept it.
            – Jon
            1 hour ago















          up vote
          0
          down vote



          accepted










          You can represent your function as
          $$
          f(x)=[theta(x+c)-theta(x-c)]sin(ax)
          $$
          being $theta(x)$ the Heaviside step function. Then, the Fourier integral becomes rather easy as
          $$
          {cal F}(f)(k)=int_{-infty}^infty dx e^{ikx}f(x)=int_{-c}^c dx e^{ikx}sin(ax).
          $$
          The latter integral is very well-known and rather easy to evaluate.






          share|cite|improve this answer





















          • Thank you for your reply! I get an integral involving sin(ac) and cos(kc),does this sound correct?
            – MathematicianP
            1 hour ago










          • Yes, it is correct.
            – Jon
            1 hour ago










          • Perfect! Thank you so much!
            – MathematicianP
            1 hour ago










          • If you like the answer, please accept it.
            – Jon
            1 hour ago













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          You can represent your function as
          $$
          f(x)=[theta(x+c)-theta(x-c)]sin(ax)
          $$
          being $theta(x)$ the Heaviside step function. Then, the Fourier integral becomes rather easy as
          $$
          {cal F}(f)(k)=int_{-infty}^infty dx e^{ikx}f(x)=int_{-c}^c dx e^{ikx}sin(ax).
          $$
          The latter integral is very well-known and rather easy to evaluate.






          share|cite|improve this answer












          You can represent your function as
          $$
          f(x)=[theta(x+c)-theta(x-c)]sin(ax)
          $$
          being $theta(x)$ the Heaviside step function. Then, the Fourier integral becomes rather easy as
          $$
          {cal F}(f)(k)=int_{-infty}^infty dx e^{ikx}f(x)=int_{-c}^c dx e^{ikx}sin(ax).
          $$
          The latter integral is very well-known and rather easy to evaluate.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Jon

          4,29511022




          4,29511022












          • Thank you for your reply! I get an integral involving sin(ac) and cos(kc),does this sound correct?
            – MathematicianP
            1 hour ago










          • Yes, it is correct.
            – Jon
            1 hour ago










          • Perfect! Thank you so much!
            – MathematicianP
            1 hour ago










          • If you like the answer, please accept it.
            – Jon
            1 hour ago


















          • Thank you for your reply! I get an integral involving sin(ac) and cos(kc),does this sound correct?
            – MathematicianP
            1 hour ago










          • Yes, it is correct.
            – Jon
            1 hour ago










          • Perfect! Thank you so much!
            – MathematicianP
            1 hour ago










          • If you like the answer, please accept it.
            – Jon
            1 hour ago
















          Thank you for your reply! I get an integral involving sin(ac) and cos(kc),does this sound correct?
          – MathematicianP
          1 hour ago




          Thank you for your reply! I get an integral involving sin(ac) and cos(kc),does this sound correct?
          – MathematicianP
          1 hour ago












          Yes, it is correct.
          – Jon
          1 hour ago




          Yes, it is correct.
          – Jon
          1 hour ago












          Perfect! Thank you so much!
          – MathematicianP
          1 hour ago




          Perfect! Thank you so much!
          – MathematicianP
          1 hour ago












          If you like the answer, please accept it.
          – Jon
          1 hour ago




          If you like the answer, please accept it.
          – Jon
          1 hour ago


















           

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