# Laplace transform of $t \cos(t)$ by definition

I want to find the Laplace transform of $t \cos(t)$ by the definition $$\int e^{-st} t \cos(2t)dt$$ The solution manual just say try the $$u = t, dv = e^{-st} \cos(2t)$$ I use the integration by parts, but still cannot delete any function. Where is the problem?

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## 2 Answers

First note that it is sufficient to find $$I(s) = \int e^{-st} \cos(2t) dt$$ since $$I'(s) = \int - t e^{-st} \cos(2t) dt$$ which gives us the desired integral we are looking for. Now note that $$I(s) = \int e^{-st} \cos(2t) dt = -\dfrac1s \int \cos(2t) d\left( e^{-st} \right)$$ Hence, $$\int \cos(2t) d\left( e^{-st} \right) = \cos(2t) e^{-st} - \int e^{-st} d\left( \cos(2t) \right) = \cos(2t) e^{-st} + 2 \int e^{-st} \sin(2t) dt$$ Now $$\int e^{-st} \sin(2t) dt = -\dfrac1s \int \sin(2t) d \left( e^{-st}\right)$$ $$\int \sin(2t) d \left( e^{-st}\right) = \sin(2t) e^{-st} - \int e^{-st} d(\sin(2t)) = \sin(2t) e^{-st} - 2 \int e^{-st} \cos(2t) dt$$ Hence, putting all this together, we get that $$I(s) = - \dfrac1s \left(\cos(2t) e^{-st}- \dfrac2s \left(\sin(2t) e^{-st} - 2 I(s) \right) \right)$$ This gives us $$s^2 I(s) = -\left(s \cos(2t) e^{-st} - 2\sin(2t)e^{-st} + 4 I(s) \right)$$ Hence, we get that $$\left(s^2 + 4\right) I(s) = \left(2 \sin(2t) - s \cos(2t) \right) e^{-st}$$ This gives us that $$I(s) = \dfrac{2 \sin(2t) - s \cos(2t)}{s^2 + 4} e^{-st} + c$$ Now $$J(s) = \int_0^{\infty} e^{-st} \cos(2t) dt = \dfrac{s}{s^2+4}$$ The Laplace transform is given by $$L(t\cos(2t)) = \int t e^{-st} \cos(2t) dt = - J'(s) = \dfrac{s^2-4}{\left(s^2+4 \right)^2}$$

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Very clear!Thx, it help me a lot!Sorry about that I cannot vote up due to my reputation is only 8. – Liang-Yu Pan Dec 16 '12 at 7:41
@Liang-YuPan No problem :). Good to know that it was helpful. – user17762 Dec 16 '12 at 7:42

Hint:
Integrate $e^{-st}\cos{2t}$ by parts twice.

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