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$$(e^{-iat-st})\over (-ja-s)$dt
It is here where I am stuck at in getting the final result.
$\endgroup$ 15 Answers
$\begingroup$You have answered it correctly, all you have to do is just pluging in the limits of integral. The first integral is \begin{align} \left.\frac{te^{-(s-ia)t}}{ia-s}\right|_0^\infty-\frac{1}{ia-s}\int_0^\infty e^{-(s-ia)t}\,dt&=0-\left.\frac{e^{-(s-ia)t}}{(ia-s)^2}\right|_0^\infty\\ &=\frac{1}{(ia-s)^2} \end{align} And the second integral is \begin{align} -\left.\frac{te^{-(s+ia)t}}{ia+s}\right|_0^\infty+\frac{1}{ia+s}\int_0^\infty e^{-(s+ia)t}\,dt&=0-\left.\frac{e^{-(s+ia)t}}{(ia+s)^2}\right|_0^\infty\\ &=\frac{1}{(ia+s)^2} \end{align} Then the answer is \begin{align} F(s)=\frac{1}{2i}\left(\frac{1}{(ia-s)^2}-\frac{1}{(ia+s)^2}\right)=\frac{2as}{\left(a^2+s^2\right)^2} \end{align}
$\endgroup$ 2 $\begingroup$I obtain the following:\begin{eqnarray*} f(t) &=&t\sin (at) \\ \int_{0}^{\infty }dtf(t)\exp [-st] &=&\int_{0}^{\infty }dtt\sin (at)\exp [-st]=-\partial _{a}\int_{0}^{\infty }dt\cos (at)\exp [-st] \\ &=&-\frac{1}{2}\partial _{a}\int_{0}^{\infty }dt\{\exp [iat]-\exp [-iat]\}\exp [-st] \\ &=&-\frac{1}{2}\partial _{a}\int_{0}^{\infty }dt\{\exp [-(s-ia)t]-\exp [-(s+ia)t]\} \\ &=&-\frac{1}{2}\partial _{a}\{\frac{1}{s-ia}+\frac{1}{s+ia}\}=-\frac{1}{2}% \partial _{a}\frac{2s}{s^{2}+a^{2}}=\frac{2as}{(s^{2}+a^{2})^{2}} \end{eqnarray*}
$\endgroup$ $\begingroup$As pointed out by Priyanka, you can twice differentiate $f(t)$ to obtain $$ f''(t) = 2acos(at) - a^2tsin(at) $$ Which is $$ f''(t) = 2acos(at) - a^2f(t) $$ Now take transform of both sides and use the fact that $$ L(f'') = s^2F(s)-sf(0)-f'(0) $$ to find the required transform.
$\endgroup$ $\begingroup$$$f′′(t)=2a\cos(at)−a2t\sin(at)$$
Which is
$$f′′(t)=2a\cos(at)−a2f(t)$$
Now take transform of both sides and use the fact that
$$L(f′′)=s2F(s)−sf(0)−f′(0)$$
to find the required transform.
$\endgroup$ $\begingroup$its easier if you try doing it by laplace transform of derivatives method
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