how you get the laplace of 1/tsin(wt)
pls show me the solution
I will leave the details of why to you.
Given $\displaystyle f(w,t)=\frac{1}{t}\sin(wt) \implies \frac{\partial f}{\partial w}=\cos(wt) $
$\displaystyle \mathcal{L}\{ \frac{\partial f}{\partial x}\}=\frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}$
$\displaystyle \frac{s}{s^2+w^2}=\frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}$
Now integrate both sides with respect to w to solve for the Laplace transform you want
$\displaystyle \int \frac{s}{s^2+w^2}dw=\int \frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}dw$
$\displaystyle \tan^{-1}\left( \frac{w}{s}\right)=\mathcal{L}\{ f(w,t)\}$
Another method is to note that the integral converges uniformly so limits and integrals can be interchanged.
Note that
$\displaystyle \sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!} \quad \tan^{-1}(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)}$
This gives that
$\displaystyle \int_{0}^{\infty}\frac{e^{-st}}{t}\left( \sum_{n=0}^{\infty}\frac{(-1)^n(\alpha t)^{2n+1}}{(2n+1)!}\right)dt = \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!}\int_{0}^{\infty} e^{-st} t^{2n}dt = \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!} \mathcal{L}\{t^{2n}\}$
Now integrate or use the definition of the Laplace transform to get
$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!}\frac{(2n)!}{s^{2n+1}}= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\left( \frac{\alpha}{s}\right)^{2n+1}=\tan^{-1}\left( \frac{\alpha}{s} \right)$