I am given the Fourier transform of $f$ by $\widehat{f}(\xi) = \frac{1}{1 + \xi^{4}}$ and have to determine the Fourier transform $\widehat{g}$ of $g(x) = f(x) \cos(2 \pi x)$.
I tried reconstructing $f$ from its Fourier transform such that $f(x) = \int_{-\infty}^{\infty} \widehat{f}(\xi)e^{2 \pi i x \xi}d\xi$ and then using $f$ to directly compute the Fourier transform of $g$ given by $\widehat{g}(\xi) = \int_{- \infty}^{\infty} g(x)e^{-2 \pi i x \xi}dx$.
But I am always stuck on these integrals. Is there perhaps some identity/trick that works around the direct computations?
Best Answer
Let $$\mathcal{F}{f}(s) = \int_{-\infty}^{+\infty}f(x)e^{-2\pi isx }dx$$We have $$\mathcal{F}(fg) = \mathcal{F}{f}*\mathcal{F}g$$Where $*$ denotes convolution. Also we have $$\mathcal{F}\cos 2\pi a t = \frac{1}{2}(\delta(s-a)+\delta(s+a))$$ Here $a = 1$ and the result is $$\mathcal{F}g(s) = (\frac{1}{2}(\delta(s-1)+\delta(s+1)))*\frac{1}{1 + s^{4}} = \frac{1}{2}(\frac{1}{1+(s-1)^4} + \frac{1}{1+(s+1)^4})$$