Answer to Question #208279 in Calculus for Abuabu

Question

Find the laplace transform of sinat where a is a constant

Answer 1

$\begin{aligned} \mathcal{L}[f(t)] &=F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t=\int_{0}^{\infty} t^{2} \sin (a t) e^{-s t} d t \\ &=\lim _{A \rightarrow \infty} \int_{0}^{A} t^{2} \sin (a t) e^{-s t} d t\\ &=\left[\begin{array}{cc} u=t & d v=t \sin (a t) e^{-s t} d t \\ d u=d t & v=\int t \sin (a t) e^{-s t} d t \end{array}\right]--(*) \end{aligned}$

$\begin{aligned} v &=\int t \sin (a t) e^{-s t} d t=\frac{1}{2 i}\left[\int t e^{(i a-s) t} d t-\int t e^{-(i a+s) t} d t\right] \\ &=\frac{1}{2 i}\left[t \frac{e^{(i a-s) t}}{i a-s}-\int \frac{e^{(i a-s) t}}{i a-s} d t+t \frac{e^{-(i a+s) t}}{i a+s}-\int \frac{e^{-(i a+s) t}}{i a+s} d t\right] \\ &=\frac{1}{2 i}\left[t \frac{e^{(i a-s) t}}{i a-s}-\frac{e^{(i a-s) t}}{(i a-s)^{2}}+t \frac{e^{-(i a+s) t}}{i a+s}+\frac{e^{-(i a+s) t}}{(i a+s)^{2}}\right] \end{aligned} --(i)$

Integrating v, we obtain;

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

$\begin{array}{l} =\frac{1}{2 i}\left\{\frac{1}{i a-s}\left[t \frac{e^{(i a-s) t}}{i a-s}-\frac{e^{(i a-s) t}}{(i a-s)^{2}}\right]-\frac{e^{(i a-s) t}}{(i a-s)^{3}}\right. \\ \left.-\frac{1}{i a+s}\left[t \frac{e^{-(i a+s) t}}{i a+s}+\frac{e^{-(i a+s) t}}{(i a+s)^{2}}\right]-\frac{e^{-(i a+s) t}}{(i a+s)^{3}}\right\} \\ =\frac{1}{2 i}\left[t \frac{e^{(i a-s) t}}{(i a-s)^{2}}-2 \frac{e^{(i a-s) t}}{(i a-s)^{3}}-t \frac{e^{-(i a+s) t}}{(i a+s)^{2}}-2 \frac{e^{-(i a+s) t}}{(i a+s)^{3}}\right] \end{array}$

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

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

$\begin{aligned} F(s) &=\frac{-2}{2 i}\left[\frac{1}{(i a-s)^{3}}+\frac{1}{(i a+s)^{3}}\right]=\frac{-1}{i} \frac{(i a+s)^{3}+(i a-s)^{3}}{[(i a-s)(i a+s)]^{3}} \\ &=\frac{-1}{i} \frac{-i a^{3}-3 a^{2} s+3 i a s^{2}+s^{3}+\left(-i a^{3}+3 a^{2} s+3 i a s^{2}-s^{3}\right)}{\left(-a^{2}-s^{2}\right)^{3}} \\ &=\frac{-1}{i} \frac{-2 i a^{3}+6 i a s^{2}}{-\left(a^{2}+s^{2}\right)^{3}}=\frac{2 a\left(3 s^{2}-a^{2}\right)}{\left(a^{2}+s^{2}\right)^{3}} \end{aligned}$