Laplace Transform of tsin(at) using only the definition Hello I' am stuck on how to get the final result of the laplace transform of $f(t)=tsin(at)$using (a is a constant) only the definition of $$\int_0^{\infty}f(t)e^{-st}dt$$,

I know $sin(at)= {1 \over 2i} (e^{iat} - e^{-iat})$

using that I get

$F(s)$=${1 \over 2i}$$[\int_0^{\infty}te^{iat-st}dt-\int_0^{\infty}te^{iat-st}dt]$

using integration by parts for each part I get

Part I

$u_1=t$

$du_1=dt$

$v_1$$=$$e^{iat-st}\over(ia-s)$

$dv_1$$=$$e^{iat-st}$

$te^{iat-st}\over(ia-s)$-$\int$$(e^{iat-st})\over (ja-s)$dt

Part 2

$u_2=t$

$du_2=dt$

$v_2$$=$$e^{-iat-st}\over(-ia-s)$

$dv_2$$=$$e^{-iat-st}$

$te^{-iat-st}\over(-ia-s)$-$\int$$(e^{-iat-st})\over (-ja-s)$dt

It is here where I am stuck at in getting the final result.
You have answered it correctly, all you have to do is just pluging in the limits of integral. The first integral is
\begin{align}
\left.\frac{te^{-(s-ia)t}}{ia-s}\right|_0^\infty-\frac{1}{ia-s}\int_0^\infty e^{-(s-ia)t}\,dt&amp;=0-\left.\frac{e^{-(s-ia)t}}{(ia-s)^2}\right|_0^\infty\\
&amp;=\frac{1}{(ia-s)^2}
\end{align}
And the second integral is
\begin{align}
-\left.\frac{te^{-(s+ia)t}}{ia+s}\right|_0^\infty+\frac{1}{ia+s}\int_0^\infty e^{-(s+ia)t}\,dt&amp;=0-\left.\frac{e^{-(s+ia)t}}{(ia+s)^2}\right|_0^\infty\\
&amp;=\frac{1}{(ia+s)^2}
\end{align}
Then the answer is
\begin{align}
F(s)=\frac{1}{2i}\left(\frac{1}{(ia-s)^2}-\frac{1}{(ia+s)^2}\right)=\frac{2as}{\left(a^2+s^2\right)^2}
\end{align}
its easier if you try doing it by laplace transform of derivatives method