A question on the Laplace Transform of $f(t)=t e^{at}\sin (bt)$
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Hint. From standard properties (see this table) and from the given result one gets $$ \mathscr{L}\left(t \sin bt \right)(s)=\left(\mathscr{L}\left( \sin bt \right)\right)'(s)=\frac{2bs}{(s^2+b^2)^2} $$ then using $$ \mathscr{L}\left(e^{at}f(t) \right)(s)=\left(\mathscr{L}f\right)(sa) $$ gives finally
$$ \mathscr{L}\left(t e^{at}\sin (bt) \right)(s)=\frac{2b(sa)}{((sa)^2+b^2)^2}. $$
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jh123
Updated on November 21, 2022Comments

jh123 12 months
I would like to solve the Laplace transform of the following function:
$$t \mapsto t e^{at}\sin (bt).$$
I know that $\mathscr{L}\left(\sin(bt)\right)=\dfrac{b}{s^2+b^2}$ and that you have to take the integral from $0$ to $\infty$ and multiply it by $e^{st}$.

Alex Provost over 7 yearsWhat is your question?

jh123 over 7 yearsJust looking to get some help with this calculation and integral

reuns over 7 yearsstart with using Latex, and not writing $\sin bt = \frac{b}{s^2+b^2}$ !


jh123 over 7 yearsOkay so then I just need to evauluate from 0 to infinity?

Olivier Oloa over 7 years@javahelper123 It's already done.

jh123 over 7 yearsIsn't e^(at) = 1/(sa)

Olivier Oloa over 7 years@javahelper123 Yes. But here we have $e^{at}\cdot f(t)$, not only $e^{at}$.

jh123 over 7 yearsAnd why do you put the s on the end in brackets like that?

Olivier Oloa over 7 yearsBecause the variable in the initial function is $t$, whereas the variable in the Laplace transform is $s$.

jh123 over 7 yearsOkay and where you say e^(at)•f(t) is f(t) e^(st)

Olivier Oloa over 7 yearsIt's rather $\mathscr{L}\left(e^{at}\cdot f(t) \right)(s)=\left(\mathscr{L}f\right)(sa)$. Thus if you know the L transform of $f$, you just deduce the L transform of $e^{at} \cdot f$ this way.

jh123 over 7 yearsOkay so this should be good enough as a solution to this question?

Olivier Oloa over 7 yearsI think it's fine as a solution. Thank you.

jh123 over 7 yearsOkay awesome thanks for your time

reuns over 7 years@javahelper123 of course if you think that $\sin b t = \frac{b^2}{b^2+s^2}$ (in latex : $\text{\$\sin b t = \frac{b^2}{b^2+s^2}\$}$) you won't go very far

Olivier Oloa over 7 years@user1952009 You are right, but I think the OP means the 'Laplace transform of...', that's why I've edited his question accordingly.

reuns over 7 yearsyes saw that, what worries me is that he didn't edit, and wrote again 'Isn't e^(at) = 1/(sa) ' (@javahelper123)