What is the result of using Euler's equation for Fourier transform integrals?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 2K views
Galizius
Messages
14
Reaction score
0
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 separate results? And if it is sum, when the imaginary or real part is being reduced to 0
 
Last edited:
Physics news on Phys.org
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].
 
  • Like
Likes   Reactions: Dr. Courtney
HallsofIvy said:
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.