# 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 $$\mathscr{L}\left(e^{at}f(t) \right)(s)=\left(\mathscr{L}f\right)(s-a)$$ gives finally

$$\mathscr{L}\left(t e^{at}\sin (bt) \right)(s)=\frac{2b(s-a)}{((s-a)^2+b^2)^2}.$$

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### jh123

Updated on November 21, 2022

• 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 years
• jh123 over 7 years
Just looking to get some help with this calculation and integral
• reuns over 7 years
start with using Latex, and not writing $\sin bt = \frac{b}{s^2+b^2}$ !
• jh123 over 7 years
Okay so then I just need to evauluate from 0 to infinity?
• Olivier Oloa over 7 years
• jh123 over 7 years
Isn't e^(at) = 1/(s-a)
• Olivier Oloa over 7 years
@javahelper123 Yes. But here we have $e^{at}\cdot f(t)$, not only $e^{at}$.
• jh123 over 7 years
And why do you put the s on the end in brackets like that?
• Olivier Oloa over 7 years
Because the variable in the initial function is $t$, whereas the variable in the Laplace transform is $s$.
• jh123 over 7 years
Okay and where you say e^(at)•f(t) is f(t) e^(-st)
• Olivier Oloa over 7 years
It's rather $\mathscr{L}\left(e^{at}\cdot f(t) \right)(s)=\left(\mathscr{L}f\right)(s-a)$. 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 years
Okay so this should be good enough as a solution to this question?
• Olivier Oloa over 7 years
I think it's fine as a solution. Thank you.
• jh123 over 7 years
Okay 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 years
yes saw that, what worries me is that he didn't edit, and wrote again 'Isn't e^(at) = 1/(s-a) ' (@javahelper123)