Euler equation

  • #1
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when I am using Euler equation for Fourier transform integrals of type [tex]\int_{-\infty}^{\infty} dx f(x) exp[ikx] [/tex]I am getting following integrals:

[tex]\int_{-\infty}^{\infty} dx f(x) cos(kx)[/tex] (for the real part) and

[tex]i* \int_{-\infty}^{\infty} dx f(x) sin(kx)[/tex] (for its imaginary part)

I am wondering what is the final integration result though. Is that the sum of both parts or are they seperate results? And if it is sum, when the imaginary or real part is being reduced to 0
 
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Answers and Replies

  • #2
Dr. Courtney
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The Fourier transform is the sum of both real and imaginary parts.
 
  • #3
HallsofIvy
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Surely if you know that [itex]e^{ikx}= cos(kx)+ i sin(kx)[/itex] then you know that [itex]\int f(x)e^{ikx}dx= \int (f(x)cos(kx)+ if(x)sin(kx))dx= \int f(x)cos(kx) dx+ i \int f(x)sin(kx) dx[/itex].
 
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  • #4
Dr. Courtney
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Surely if you know that [itex]e^{ikx}= cos(kx)+ i sin(kx)[/itex] then you know that [itex]\int f(x)e^{ikx}dx= \int (f(x)cos(kx)+ if(x)sin(kx))dx= \int f(x)cos(kx) dx+ i \int f(x)sin(kx) dx[/itex].

Well, when you put it that way ...

Nice proof. Thanks.
 
  • #5
what is the complete form of euler equation?
 

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