Question Number 147711 by lapache last updated on 22/Jul/21
$${Determiner}\:{l}'{original}\:{de}\:{laplace} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by Olaf_Thorendsen last updated on 23/Jul/21
![G(x) = F(x−(1/2)) = (1/((x^2 +(3/4))^2 )) H(x) = G(((√3)/2)x) = ((16)/9).(1/((x^2 +1)^2 )) H(x) = ((16)/9)[−(1/(4(x+i)^2 ))−(1/(4(x−i)^2 ))+(i/(4(x+i)))−(i/(4(x−i)))] H(x) = (4/9)[−(1/((x+i)^2 ))−(1/((x−i)^2 ))+(i/(x+i))−(i/(x−i))] L^(−1) ((1/(x−a))) = e^(at) et L^(−1) ((1/((x−a)^2 ))) = te^(at) L^(−1) {H}(t) = (4/9)[−te^(−it) −te^(it) +ie^(−it) −ie^(it) ] L^(−1) {H}(t) = (4/9)[−2tcost+2sint] L^(−1) {H}(t) = (8/9)(sint−tcost) L^(−1) {G(x)}(t) = L^(−1) {H((x/((√3)/2)))}(t) L^(−1) {G(x)}(t) = ((√3)/2)((1/((√3)/2))L^(−1) {H((x/((√3)/2)))}(t) ) L^(−1) {G(x)}(t) = ((√3)/2)(L^(−1) {H}(((√3)/2)t)) L^(−1) {G}(t) = (4/(3(√3)))(sin(((√3)/2)t)−((√3)/2)tcos(((√3)/2)t)) L^(−1) {F(x)}(t) = L^(−1) {G(x+(1/2))}(t) L^(−1) {F(x)}(t) = e^(−(1/2)t) L^(−1) {G(x)}(t) L^(−1) {F}(t) = (4/(3(√3)))e^(−(t/2)) (sin(((√3)/2)t)−((√3)/2)tcos(((√3)/2)t))](Q147723.png)
$$\mathrm{G}\left({x}\right)\:=\:{F}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$${H}\left({x}\right)\:=\:{G}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right)\:=\:\frac{\mathrm{16}}{\mathrm{9}}.\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${H}\left({x}\right)\:=\:\frac{\mathrm{16}}{\mathrm{9}}\left[−\frac{\mathrm{1}}{\mathrm{4}\left({x}+{i}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}\left({x}−{i}\right)^{\mathrm{2}} }+\frac{{i}}{\mathrm{4}\left({x}+{i}\right)}−\frac{{i}}{\mathrm{4}\left({x}−{i}\right)}\right] \\ $$$${H}\left({x}\right)\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[−\frac{\mathrm{1}}{\left({x}+{i}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({x}−{i}\right)^{\mathrm{2}} }+\frac{{i}}{{x}+{i}}−\frac{{i}}{{x}−{i}}\right] \\ $$$$\mathcal{L}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}−{a}}\right)\:=\:{e}^{{at}} \:\mathrm{et}\:\mathcal{L}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\left({x}−{a}\right)^{\mathrm{2}} }\right)\:=\:{te}^{{at}} \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{{H}\right\}\left({t}\right)\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[−{te}^{−{it}} −{te}^{{it}} +{ie}^{−{it}} −{ie}^{{it}} \right] \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{{H}\right\}\left({t}\right)\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[−\mathrm{2}{t}\mathrm{cos}{t}+\mathrm{2sin}{t}\right] \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{{H}\right\}\left({t}\right)\:=\:\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{sin}{t}−{t}\mathrm{cos}{t}\right) \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{\mathrm{G}\left({x}\right)\right\}\left({t}\right)\:=\:\mathcal{L}^{−\mathrm{1}} \left\{\mathrm{H}\left(\frac{{x}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)\right\}\left({t}\right)\: \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{\mathrm{G}\left({x}\right)\right\}\left({t}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\mathcal{L}^{−\mathrm{1}} \left\{\mathrm{H}\left(\frac{{x}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)\right\}\left({t}\right)\:\right) \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{\mathrm{G}\left({x}\right)\right\}\left({t}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathcal{L}^{−\mathrm{1}} \left\{\mathrm{H}\right\}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\right)\right) \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{\mathrm{G}\right\}\left({t}\right)\:=\:\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}\left(\mathrm{sin}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\right)\right) \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{{F}\left({x}\right)\right\}\left({t}\right)\:=\:\mathcal{L}^{−\mathrm{1}} \left\{{G}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}\left({t}\right) \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{{F}\left({x}\right)\right\}\left({t}\right)\:=\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{t}} \mathcal{L}^{−\mathrm{1}} \left\{{G}\left({x}\right)\right\}\left({t}\right) \\ $$$$ \\ $$$$\mathcal{L}^{−\mathrm{1}} \left\{{F}\right\}\left({t}\right)\:=\:\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}{e}^{−\frac{{t}}{\mathrm{2}}} \left(\mathrm{sin}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\right)\right) \\ $$