0
521views
Using Laplace Transform evaluate \[\int_0^{\infty{}}e^{-t}\frac{\sin{3t}}{t}dt.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \]
1 Answer
0
3views

$Consider,\\[3ex] \displaystyle \ L\left[sin3t\right]=\frac{3}{s^2+3^2} \\[3ex] \displaystyle \therefore{}L\left[\frac{\sin3t}{t}\right]\\[2ex] \displaystyle =\int_s^{\infty{}}\frac{3}{s^2+3^2}ds \\[2ex] \displaystyle =3\times{}\frac{1}{3}\times{}{\left[{\tan}^{-1}{\frac{s}{3}}\right]}_s^{\infty{}} \\[2ex] \displaystyle =\frac{\pi{}}{2}-{\tan}^{-1}{\frac{s}{3}} \\[2ex] $

$\displaystyle \therefore{}\ \int_0^{\infty{}}e^{-t}\frac{\sin{3t}}{t}dt={\cot}^{-1}{\frac{s}{3}}$

$ Put\ s=1\\[3ex] \displaystyle \therefore{}\ \int_0^{\infty{}}e^{-t}\frac{\sin{3t}}{t}dt={\cot}^{-1}{\frac{s}{3}}$

Please log in to add an answer.