Fourier transform of $\cos(kx)$ using the formula given.

Question

I want to find the Fourier transform of $$ f(x) = \cos (kx)$$ using the Fourier transformation formula $$f(k)={1\over \sqrt{2\pi}}\int _{-\infty}^\infty f(x) e^{ikx}dk$$

How can I do that?

Answer 1

As tired mentioned, a delta function, $\delta(x)$, is a distribution that satisfies $$\int _{-\infty}^\infty\delta(x-a)f(x)=f(a)$$ Indeed the Fourier transform of the function $g(x)=1$ is a delta function.So we should Write $$\delta(\omega)={1\over \sqrt{2\pi}}\int _{-\infty}^\infty e^{i\omega x}dx$$ $$\mathcal{F}[f(x)](\omega)={1\over \sqrt{2\pi}}\int _{-\infty}^\infty \cos (kx)\,e^{i\omega x}dx={1\over \sqrt{2\pi}}\int _{-\infty}^\infty \left(\frac{e^{ikx}+e^{-ikx}}{2}\right) e^{i\omega x}dx$$ we have $$\mathcal{F}[f(x)](\omega)={1\over 2\sqrt{2\pi}}\int _{-\infty}^\infty \left(e^{ix(\omega +k)}+e^{ix(\omega -k)}\right) dx$$ $$\mathcal{F}[f(x)](\omega)={1\over 2}\delta(\omega+k)+\frac 12\delta(\omega-k)$$