The Fourier transform of a function that is spherically symmetric $g(r)$ is equal to \begin{equation} \tilde h = \frac{4\pi}{k} \int_0^\infty \text{d}r\: \sin(kr) rg(r) \end{equation} (https://www.theoretical-physics.net/dev/math/transforms.html). However, in their book Physical Chemistry – An Advanced Treatise, Vol. VIIIA Eyring, Henderson and Jost state that applying integration by parts to the previous expression gives \begin{equation} \tilde h = 4\pi \int_0^\infty \text{d}r\: (\cos(kr))G(r) , \end{equation} where $G(r)=\int_r^\infty \text{d}t\:tg(t)$. I am struggling to derive this equation. I suspect that both forms of the Fourier transform is common knowledge for people who use Fourier transforms regularly.
Any help with deriving the second expression from the first is appreciated.
Let $G(r) = \int_r^\infty dt \, t g(t).$ Then, $G'(r) = -r g(r)$ so $$ \int_0^\infty dr \sin(kr) \, r g(r) = -\int_0^\infty dr \sin(kr) \, G'(r) = -\left( \left[ \sin(kr) \, G(r) \right]_0^\infty - \int_0^\infty dr \, k\cos(kr) \, G(r) \right) \\ = k \int_0^\infty dr \cos(kr) \, G(r) - \left[ \sin(kr) \, G(r) \right]_0^\infty . $$ Assuming that $g(r)$ falls off fast enough so that $G(r) \to 0$ as $r\to\infty$ the last term vanishes.