if $k(x)$ is its own Fourier Cosine Transform, show that $g(x)=\int_0^\infty k(t)e^{-xt}dt $ is its own Fourier Sine Transform.
I have no idea how to go about this. I think:
since $k(x)$ is its own FCT, then $g(x)=\int_0^\infty \int_0^\infty k(x)\cos(xt) dx e^{-xt}dt $ but I am not sure
Let be $$ \mathcal F_c\{k(t)\}=\sqrt\frac{2}{\pi}\int_0^\infty k(t)\cos(\omega t)\,\mathrm dt=k(\omega) $$ the Fourier cosine transform of $k$ and $$ \mathcal{L}\{k(t)\}=\int_0^\infty k(t)\,e^{-x t}\,\mathrm dt=g(x) $$ the Laplace transform of $k$. So for the Fourier sine transform of $g(x)$ we have $$ \mathcal F_s\{g(x)\}=\sqrt\frac{2}{\pi}\int_0^\infty g(x)\sin(\xi x)\,\mathrm d x=g(\xi) $$
In fact $$\begin{align} \mathcal F_s\{g(x)\}&=\sqrt\frac{2}{\pi}\int_0^\infty \color{blue}{g(x)}\sin(\xi x)\,\mathrm dx\\ &=\sqrt\frac{2}{\pi}\int_0^\infty \color{blue}{\int_0^\infty} \color{red}{k(t)}\color{blue}{\,e^{-x t}\,\mathrm dt}\sin(\xi x)\,\mathrm dx\\ &=\sqrt\frac{2}{\pi}\int_0^\infty \color{blue}{\int_0^\infty} \underbrace{\color{red}{\sqrt\frac{2}{\pi}\int_0^\infty k(\omega)\cos(\omega t)\,\mathrm d\omega}}_{\mathcal F_c^{-1}\{k(\omega)\}}\color{blue}{\,e^{-x t}\,\mathrm dt}\sin(\xi x)\,\mathrm dx\\ &=\sqrt\frac{2}{\pi}\int_0^\infty \color{red}{\int_0^\infty} \color{red} {k(\omega)}\underbrace{\color{blue}{\sqrt\frac{2}{\pi}\int_0^\infty\cos(\omega t)}\color{blue}{\,e^{-x t}\,\mathrm dt}}_{\mathcal F_c\{e^{-xt}\}} \color{red} {\,\mathrm d\omega} \sin(\xi x)\,\mathrm dx\\ &=\sqrt\frac{2}{\pi}\int_0^\infty \color{red} {k(\omega)}\underbrace {\color{green}{\sqrt\frac{2}{\pi}\int_0^\infty \frac{x}{x^2+\omega^2} \sin(\xi x)\,\mathrm dx}}_{\mathcal F_s\left\{ \frac{x}{x^2+\omega^2} \right\}}\color{red} {\,\mathrm d\omega}\\ &=\sqrt\frac{2}{\pi}\int_0^\infty \color{red} {k(\omega)}\color{green}{\sqrt\frac{\pi}{2}e^{-\omega \xi}}\color{red} {\,\mathrm d\omega}\\ &=\int_0^\infty k(\omega){e^{-\omega \xi}} {\,\mathrm d\omega}\\ &=g(\xi) \end{align} $$