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?

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

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

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


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