Answer to Question #228608 in Math for abi

Question

L -1 {k / s²+k²} =sinkt

Answer 1

$$L(\sin(kt))=\displaystyle\int_{0}^{\infin}e^{-ts}\sin(kt)dt$$

$$\int e^{-ts}\sin(kt)dt=-\dfrac{1}{s}e^{-ts}\sin(kt)$$

$$+\dfrac{k}{s}\int e^{-ts}\cos(kt)dt$$

$$=-\dfrac{1}{s}e^{-ts}\sin(kt)-\dfrac{k}{s^2}e^{-ts}\cos(kt)$$

$$-\dfrac{k^2}{s^2}\int e^{-ts}\sin(kt)dt$$

Then

$$(s^2+k^2)\int e^{-ts}\sin(kt)dt=-e^{-ts}(s\sin(kt)+k\cos(kt)+C_1$$

$$\int e^{-ts}\sin(kt)dt=-\dfrac{e^{-ts}(s\sin(kt)+k\cos(kt)}{s^2+k^2}+C$$

$$L(\sin(kt))=\displaystyle\int_{0}^{\infin}e^{-ts}\sin(kt)dt$$

$$=\lim\limits_{A\to \infin}[-\dfrac{e^{-ts}(s\sin(kt)+k\cos(kt)}{s^2+k^2}]\begin{matrix} A \\ 0 \end{matrix}$$

$$=0+\dfrac{k}{s^2+k^2}$$

$$L(\sin(kt))=\dfrac{k}{s^2+k^2}$$

$$L^{-1}(\dfrac{k}{s^2+k^2})=\sin(kt)$$