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

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Using Euler's equation for Fourier transform integrals results in separating the integral into real and imaginary parts: the real part corresponds to the integral of the cosine function, while the imaginary part involves the sine function. The final integration result is the sum of both parts, expressed as the integral of the cosine plus an imaginary component from the sine integral. The relationship e^(ikx) = cos(kx) + i sin(kx) confirms this separation. The discussion also touches on the need for clarity regarding the complete form of Euler's equation, directing users to additional resources for further information. Understanding these components is essential for accurately applying Fourier transforms in analysis.
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when I am using Euler equation for Fourier transform integrals of type \int_{-\infty}^{\infty} dx f(x) exp[ikx]I am getting following integrals:

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

i* \int_{-\infty}^{\infty} dx f(x) sin(kx) (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
 
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The Fourier transform is the sum of both real and imaginary parts.
 
Surely if you know that e^{ikx}= cos(kx)+ i sin(kx) then you know that \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.
 
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HallsofIvy said:
Surely if you know that e^{ikx}= cos(kx)+ i sin(kx) then you know that \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.

Well, when you put it that way ...

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