Question Number 129048 by bramlexs22 last updated on 12/Jan/21
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Laplace}\:\mathrm{transform} \\ $$$$\mathrm{of}\:\mathrm{f}\left(\mathrm{t}\right)\:=\:−\mathrm{4t}^{\mathrm{2}} −\mathrm{5sin}\:\mathrm{3t}\: \\ $$
Answered by Dwaipayan Shikari last updated on 12/Jan/21
$$\mathscr{L}\left({f}\left({t}\right)\right)=−\mathrm{4}\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−{st}} −\mathrm{5}\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} {sin}\mathrm{3}{t} \\ $$$$=\:\:−\frac{\mathrm{4}}{{s}^{\mathrm{3}} }\:\int_{\mathrm{0}} ^{\infty} {j}^{\mathrm{2}} {e}^{−{j}} {dj}−\frac{\mathrm{5}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {e}^{−\left({s}−\mathrm{3}{i}\right)} −{e}^{−\left({s}+\mathrm{3}{i}\right)} {dt} \\ $$$$=−\frac{\mathrm{8}}{{s}^{\mathrm{3}} }−\frac{\mathrm{5}}{\mathrm{2}{i}}\left(\frac{\mathrm{1}}{{s}−\mathrm{3}{i}}−\frac{\mathrm{1}}{{s}+\mathrm{3}{i}}\right)=−\frac{\mathrm{8}}{{s}^{\mathrm{3}} }−\frac{\mathrm{15}}{{s}^{\mathrm{2}} +\mathrm{9}} \\ $$