# Euler equation

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 seperate results? And if it is sum, when the imaginary or real part is being reduced to 0

Last edited:

Dr. Courtney
Gold Member
The Fourier transform is the sum of both real and imaginary parts.

HallsofIvy
Homework Helper
Surely if you know that $e^{ikx}= cos(kx)+ i sin(kx)$ then you know that $\int f(x)e^{ikx}dx= \int (f(x)cos(kx)+ if(x)sin(kx))dx= \int f(x)cos(kx) dx+ i \int f(x)sin(kx) dx$.

Dr. Courtney
Dr. Courtney
Surely if you know that $e^{ikx}= cos(kx)+ i sin(kx)$ then you know that $\int f(x)e^{ikx}dx= \int (f(x)cos(kx)+ if(x)sin(kx))dx= \int f(x)cos(kx) dx+ i \int f(x)sin(kx) dx$.