Chapter 4.12, Problem 15P

Given that $\begin{array}{l}\hfill \\ {\displaystyle {\int }_0^{\infty }{e}^{-ax}}\sin kxdx=\frac{k}{a^2+k^2}\hfill \end{array}$ differentiate with respect to $a$ to show that ${\displaystyle {\int }_0^{\infty }x}{e}^{-ax}\sin kxdx=\frac{2ka}{{\left(a^2+k^2\right)}^2}$ and differentiate with respect to $k$ to show that ${\displaystyle {\int }_0^{\infty }x}{e}^{-ax}\cos kxdx=\frac{a^2-k^2}{{\left(a^2+k^2\right)}^2}$.