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I am trying to use the convolution theorem to find the inverse Laplace transform of

$$\dfrac{s}{(s - 2)^{3/2}(s^2 + 1)}$$

The convolution theorem states that

$$f * g = \int_0^t f(\tau)g(t - \tau) \ d \tau$$

So we have

$$\dfrac{s}{s^2 + 1} = \mathcal{L}\{ \cos(t) \}$$

and

$$\dfrac{1}{(s - 2)^{3/2}} =$$

I can't figure out the last one.

I would greatly appreciate it if someone could please help me with this.

EDIT:

Ok, I got that it is $\dfrac{1}{(s - 2)^{3/2}} = \mathcal{L} \left\{ e^{2t}\sqrt{t} \dfrac{2}{\pi} \right\}$. I'm having trouble doing the integration by parts for $\int_0^t [\cos(\tau)] \left[ \dfrac{2}{\pi}e^{2t} \sqrt{t} - \tau \right] \ d \tau$. Can someone please show me how this is done?

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  • $\begingroup$ See second answer at quora.com/How-do-I-find-the-Laplace-transform-of-sqrt-t-e-3t $\endgroup$ – Moo Sep 17 '18 at 12:18
  • $\begingroup$ for "the last one" look at $\sqrt{t} e^{2t}$ ... $\endgroup$ – Math-fun Sep 17 '18 at 12:18
  • $\begingroup$ Ok, I got that it is $\dfrac{1}{(s - 2)^{3/2}} = \mathcal{L} \left\{ e^{2t}\sqrt{t} \dfrac{2}{\pi} \right\}$. I'm having trouble doing the integration by parts for $\int_0^t [\cos(\tau)] \left[ \dfrac{2}{\pi}e^{2t} \sqrt{t} - \tau \right] \ d \tau$. Can someone please show me how this is done? $\endgroup$ – The Pointer Sep 17 '18 at 13:36
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Hint:

$$\mathcal{L}[e^{at}t^n]= \frac{\Gamma (n+1)}{(s-a)^{n+1}}$$

Where $\Gamma$ is the gamma function.

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