F(s)=∫0∞e−t⋅s⋅f(t) dt==∫0∞e−t⋅s⋅t2⋅sinh(at) dt==∫0∞e−t⋅s⋅t2⋅eat−e−at2 dt==12⋅∫0∞t2⋅(e(−s+a)t−e(−s−a)t) dt==12⋅∫0∞t2⋅e−(s−a)t dt−12⋅∫0∞t2⋅e−(s+a)t dt==12⋅1(s−a)3⋅Γ(3)−12⋅1(s+a)3⋅Γ(3)=={Γ(3)=(3−1)!=2}=1(s−a)3−1(s+a)3;F(s)=\int\limits_0^\infty e^{-t\cdot s}\cdot f(t)\,dt=\\ =\int\limits_0^\infty e^{-t\cdot s}\cdot t^2\cdot\sinh(at)\,dt=\\ =\int\limits_0^\infty e^{-t\cdot s}\cdot t^2\cdot\frac{e^{at}-e^{-at}}{2}\,dt=\\ =\frac{1}{2}\cdot\int\limits_0^\infty t^2\cdot(e^{(-s+a)t}-e^{(-s-a)t})\,dt=\\ =\frac{1}{2}\cdot\int\limits_0^\infty t^2\cdot e^{-(s-a)t}\,dt-\frac{1}{2}\cdot\int\limits_0^\infty t^2\cdot e^{-(s+a)t}\,dt=\\ =\frac{1}{2}\cdot \frac{1}{(s-a)^3}\cdot\varGamma(3) - \frac{1}{2}\cdot \frac{1}{(s+a)^3}\cdot\varGamma(3)=\\ =\{\varGamma(3)=(3-1)!=2\}=\frac{1}{(s-a)^3}-\frac{1}{(s+a)^3};F(s)=0∫∞e−t⋅s⋅f(t)dt==0∫∞e−t⋅s⋅t2⋅sinh(at)dt==0∫∞e−t⋅s⋅t2⋅2eat−e−atdt==21⋅0∫∞t2⋅(e(−s+a)t−e(−s−a)t)dt==21⋅0∫∞t2⋅e−(s−a)tdt−21⋅0∫∞t2⋅e−(s+a)tdt==21⋅(s−a)31⋅Γ(3)−21⋅(s+a)31⋅Γ(3)=={Γ(3)=(3−1)!=2}=(s−a)31−(s+a)31;
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