# Answer in Calculus for Abuabu #208279

Find the laplace transform of sinat where a is a constant

“begin{aligned}nmathcal{L}[f(t)] &=F(s)=int_{0}^{infty} f(t) e^{-s t} d t=int_{0}^{infty} t^{2} sin (a t) e^{-s t} d t \n&=lim _{A rightarrow infty} int_{0}^{A} t^{2} sin (a t) e^{-s t} d t\n&=left[begin{array}{cc}nu=t & d v=t sin (a t) e^{-s t} d t \nd u=d t & v=int t sin (a t) e^{-s t} d tnend{array}right]–(*)nend{aligned}”

“begin{aligned}nv &=int t sin (a t) e^{-s t} d t=frac{1}{2 i}left[int t e^{(i a-s) t} d t-int t e^{-(i a+s) t} d tright] \n&=frac{1}{2 i}left[t frac{e^{(i a-s) t}}{i a-s}-int frac{e^{(i a-s) t}}{i a-s} d t+t frac{e^{-(i a+s) t}}{i a+s}-int frac{e^{-(i a+s) t}}{i a+s} d tright] \n&=frac{1}{2 i}left[t frac{e^{(i a-s) t}}{i a-s}-frac{e^{(i a-s) t}}{(i a-s)^{2}}+t frac{e^{-(i a+s) t}}{i a+s}+frac{e^{-(i a+s) t}}{(i a+s)^{2}}right]nend{aligned} –(i)”

Integrating v, we obtain;

“int v d t=frac{1}{2 i}left[int t frac{e^{(i a-s) t}}{i a-s} d t-frac{e^{(i a-s) t}}{(i a-s)^{3}}+int t frac{e^{-(i a+s) t}}{i a+s} d t-frac{e^{-(i a+s) t}}{(i a+s)^{3}}right]–(ii)”

“begin{array}{l}n=frac{1}{2 i}left{frac{1}{i a-s}left[t frac{e^{(i a-s) t}}{i a-s}-frac{e^{(i a-s) t}}{(i a-s)^{2}}right]-frac{e^{(i a-s) t}}{(i a-s)^{3}}right. \nleft.-frac{1}{i a+s}left[t frac{e^{-(i a+s) t}}{i a+s}+frac{e^{-(i a+s) t}}{(i a+s)^{2}}right]-frac{e^{-(i a+s) t}}{(i a+s)^{3}}right} \n=frac{1}{2 i}left[t frac{e^{(i a-s) t}}{(i a-s)^{2}}-2 frac{e^{(i a-s) t}}{(i a-s)^{3}}-t frac{e^{-(i a+s) t}}{(i a+s)^{2}}-2 frac{e^{-(i a+s) t}}{(i a+s)^{3}}right]nend{array}”

Returning back to the start and using 1 and 2, we have

“begin{aligned}& lim _{A rightarrow infty}left(left.u vright|_{0} ^{A}-int_{0}^{A} v d uright) \n=& lim _{A rightarrow infty}left{left.frac{t}{2 i}left[t frac{e^{(i a-s) t}}{i a-s}-frac{e^{(i a-s) t}}{(i a-s)^{2}}+t frac{e^{-(i a+s) t}}{i a+s}+frac{e^{-(i a+s) t}}{(i a+s)^{2}}right]right|_{0} ^{A}right.\n&left.-left.frac{1}{2 i}left[t frac{e^{(i a-s) t}}{(i a-s)^{2}}-2 frac{e^{(i a-s) t}}{(i a-s)^{3}}-t frac{e^{-(i a+s) t}}{(i a+s)^{2}}-2 frac{e^{-(i a+s) t}}{(i a+s)^{3}}right]right|_{0} ^{A}right} \n=& lim _{A rightarrow infty}left{frac{A}{2 i}left[frac{A e^{(i a-s) A}}{i a-s}-frac{e^{(i a-s) A}}{(i a-s)^{2}}+frac{A e^{-(i a+s) A}}{i a+s}+frac{e^{-(i a+s) A}}{(i a+s)^{2}}right]right.\n&-frac{1}{2 i}left[frac{A e^{(i a-s) A}}{(i a-s)^{2}}-2 frac{e^{(i a-s) A}}{(i a-s)^{3}}-frac{A e^{-(i a+s) A}}{(i a+s)^{2}}-2 frac{e^{-(i a+s) A}}{(i a+s)^{3}}right] \n&left.+frac{1}{2 i}left[-2 frac{1}{(i a-s)^{3}}-2 frac{1}{(i a+s)^{3}}right]right}nend{aligned}”

“begin{aligned}nF(s) &=frac{-2}{2 i}left[frac{1}{(i a-s)^{3}}+frac{1}{(i a+s)^{3}}right]=frac{-1}{i} frac{(i a+s)^{3}+(i a-s)^{3}}{[(i a-s)(i a+s)]^{3}} \n&=frac{-1}{i} frac{-i a^{3}-3 a^{2} s+3 i a s^{2}+s^{3}+left(-i a^{3}+3 a^{2} s+3 i a s^{2}-s^{3}right)}{left(-a^{2}-s^{2}right)^{3}} \n&=frac{-1}{i} frac{-2 i a^{3}+6 i a s^{2}}{-left(a^{2}+s^{2}right)^{3}}=frac{2 aleft(3 s^{2}-a^{2}right)}{left(a^{2}+s^{2}right)^{3}}nend{aligned}”