Sine and cosine transforms

Remarks

Both integrals converge absolutely because $\abs{\cos}$ and $\abs{\sin}$ are at most $1$. Example: for $f(x) = e^{-x}$, two integrations by parts give $\mathcal{F}_c f(\xi) = \frac{1}{1 + a^2}$ and $\mathcal{F}_s f(\xi) = \frac{a}{1 + a^2}$ with $a = 2\pi \xi$. Non-example: the constant function $f = 1$ is not integrable on $[0, \infty)$, and $\int_0^\infty \cos(2\pi x \xi)\, dx$ diverges for every $\xi$.

Let $f_e$ and $f_o$ be the even and the odd extensions of $f$ to $\R$, and let $\hat g$ denote the Fourier transform $\int_\R g(x) e^{-2\pi i x \xi}\, dx$. Then $\hat f_e(\xi) = 2 \mathcal{F}_c f(\xi)$ and $\hat f_o(\xi) = -2i\, \mathcal{F}_s f(\xi)$. Consequently, if $f$ is continuous and $\mathcal{F}_c f$ is integrable, then $f(x) = 4 \int_0^\infty \mathcal{F}_c f(\xi) \cos(2\pi x \xi)\, d\xi$ for $x \ge 0$.

Other conventions use $\cos(x \xi)$ and $\sin(x \xi)$, often with a factor $\sqrt{2/\pi}$ that makes the transforms their own inverses on $L^2(0, \infty)$.