1. We are asked to find the Fourier cosine and sine transforms of the function $f(x) = e^{-kx}$ with $k > 0$ and $x > 0$. 2. The Fourier cosine transform $F_c(w)$ is defined as: $$F_c(w) = \sqrt{\frac{2}{\pi}} \int_0^{\infty} f(x) \cos(wx) \, dx$$ Substitute $f(x) = e^{-kx}$: $$F_c(w) = \sqrt{\frac{2}{\pi}} \int_0^{\infty} e^{-kx} \cos(wx) \, dx$$ 3. Using the integral formula: $$\int_0^{\infty} e^{-ax} \cos(bx) \, dx = \frac{a}{a^2 + b^2}, \quad a > 0$$ we get: $$F_c(w) = \sqrt{\frac{2}{\pi}} \frac{k}{k^2 + w^2}$$ 4. The Fourier sine transform $F_s(w)$ is defined as: $$F_s(w) = \sqrt{\frac{2}{\pi}} \int_0^{\infty} f(x) \sin(wx) \, dx$$ Substitute $f(x) = e^{-kx}$: $$F_s(w) = \sqrt{\frac{2}{\pi}} \int_0^{\infty} e^{-kx} \sin(wx) \, dx$$ 5. Using the integral formula: $$\int_0^{\infty} e^{-ax} \sin(bx) \, dx = \frac{b}{a^2 + b^2}, \quad a > 0$$ we get: $$F_s(w) = \sqrt{\frac{2}{\pi}} \frac{w}{k^2 + w^2}$$ **Final answers:** $$F_c(w) = \sqrt{\frac{2}{\pi}} \frac{k}{k^2 + w^2}$$ $$F_s(w) = \sqrt{\frac{2}{\pi}} \frac{w}{k^2 + w^2}$$