Determine Fourier transform 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?
 A: 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 $$\mathcal{F}g(s) = (\frac{1}{2}(\delta(s-1)+\delta(s+1)))*\frac{1}{1 + s^{4}} = \frac{1}{2}(\frac{1}{1+(s-1)^4} + \frac{1}{1+(s+1)^4})$$
A: Before some comments adding some background to @S.H.W.'s very efficient approach, it may be worthwhile to give an (iconic!) argument that is easier to justify:
Since cosine is a linear combination of complex exponentials, it suffices to evaluate $\int_{-\infty}^\infty { e^{itx}\over 1+x^4 }dx$ for real $t$. There are two cases, depending on the sign of $t$. For $t\ge 0$, the function $z\to e^{itz}$ is bounded in the upper half-plane. Thus, as $R\to +\infty$, the integral of $e^{itz}/(1+z^4)$ over a semi-circle in the upper half-plane, of radius $R$, goes to $0$, by easy estimates. The integral over that auxiliary arc, together with the integral along $[-R,R]$ on the real line, is $2\pi i$ times the sum of residues inside the resulting closed contour. These occur exactly at $z=e^{2\pi i/8}$ and at $z=e^{2\pi i\cdot 3/8}$...
For $t\le 0$, we must use an arc in the lower half-plane, instead, because that's where $z\to e^{itz}$ is bounded, so that the integral over the auxiliary arc goes to $0$, and the residue theorem is easily applied.
The conversion of products to convolutions by Fourier transform is often a very good heuristic, insofar as it packages up certain standard computations usefully. But, yes, there are some hazards, especially in delineating the limits of applicability of the ideas. To begin with, yes, how to compute the Fourier transforms of functions like $e^{iz}$ or $\cos(z)$, when the integrals don't converge at all? Well, an extended sense does (provably) give self-consistent answers (as used in @S.H.W.'s discussion), and sometimes gives a more economical computational route.
Some of the potential hazards include computation of convolutions for not-classical, meaning not-pointwise-defined, functions, since the integrals cannot be literal integrals. Another hazard is about associativity of convolution: it definitely fails even in mundane-seeming situations, such as the iconic
$$
(1*\delta')*H \;=\; 0 * H \;=\; 0\;\not=\; 1\;=\; 1 * \delta \;=\; 1 * (\delta' * H) 
$$
where $1$ is the identically-one function and $H$ is Heaviside's step function.
EDIT: a non-exhaustive list of situations in which Fourier transform converts products to convolutions:
For Schwartz functions (mapped to themselves by Fourier transform), this property holds, and we have associativity. Beyond this case, things cannot remain entirely symmetrical. For example, in the iconic counter-example, all the distributions are tempered, so they have Fourier transforms. An obstacle we might anticipate is that if $f,g$ cannot be pointwise-multiplied, which could occur in part because they don't have pointwise values, then $fg$ might not be anything we could take a Fourier transform of in the first place.
Another issue which show some limitations is extension of the definition of convolution (not just "a definition", but interacting reasonably with other operations). And it can't be entirely symmetric, in light of the iconic counter-example. One valid extension is to have compactly-supported distributions $u$ act on smooth functions $f$ by $(u*f)(x)=u(T_xf)$ where $T_x$ is translation by $x$. This does indeed give another smooth function as outcome. Then there is provably a "convolution" of compactly-supported distributions such that for all smooth $f$ we have $(u*v)*f=u*(v*f)$. In fact, it might be better to use a different notation for the action, to illuminate the lack of symmetry: write $u\cdot f$... then $(u*v)\cdot f=u\cdot (v\cdot f)$.
From Paley-Wiener-type results, we know that certain entire functions' Fourier transforms are compactly-supported distributions. More typically, we start with very tangible compactly-supported distributions, and have some convenient ad-hoc way to compute their Fourier transforms (other than the obvious integral, which typically would not make sense.)
So, as some classical sources indicate, one constraint in wanting associativity of convolution is that at least two of the involved items should be compactly supported. This seeming asymmetry is genuine: as in the previous paragraph, "in reality" compactly-supported things act on not-compactly-supported, in many cases where not-compactly-supported things do not reasonably act.
