# Fourier transform of $f(x)cos(ax)$

I am trying to calculate the Fourier transform of the function $f(x)cos(ax)$ . I tried to take it by parts, putting $cos(ax)$ inside dx, but I get terrible integrals and it seems to me that this is not the right way to do it... Is there any easier method to calculate it?

• what definition of a fourier transform are you using? i.e. what is the integral you calculated? – John Doe Jan 22 '18 at 23:24
• $\cos(ax) = \frac{e^{iax} + e^{-iax}}{2}$ and the term can be absorbed into the $e^{-i\xi x}$ of the Fourier transform. – user335907 Jan 22 '18 at 23:26

Irrelevant of whatever constants you may have out front (perhaps $1/{\sqrt{2\pi}}$),

\begin{align}\mathcal F[f(x)\cos ax]&=\int f(x) \cos(ax)e^{-ikx}\,dx\\&=\frac12\int f(x)\left(e^{-i(k-a)x}+e^{-i(k+a)x}\right)\,dx\\&=\frac12\left(\tilde f(k-a)+\tilde f(k+a)\right)\end{align}