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Using the rule: $$\mathcal{L}(t^nf(t))=(-1)^n\frac{d^n}{ds^n}F(s)$$

where in this case $$f(t)=\sin(t),\,\,\,\,\,\,\,\,\,\mathcal{L}(\sin(t))=F(s)=\frac1{s^2+1},\,\,\,\,\,\,\,\,\,\,n=2.$$ Find the 2nd derivative of F(s):

$$\frac{d^2}{ds^2}\Big(\frac1{s^2+1}\Big)=\frac{6s^2-2}{(s^2+1)^3}$$

The transform:

$$\mathcal{L}(t^2sin(t))=(-1)^2\frac{6s^2-2}{(s^2+1)^3}$$ $$= \frac{6s^2-2}{(s^2+1)^3}$$

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    @ stevesy $L[f\cdot g]\neq L[f] \cdot L[g]$ – 2017-02-22
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    Which means I cannot multiply one transform by the other, but I can multiply one function by the other and then transform the result? – 2017-02-23

3 Answers 3

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$$L[f\cdot g]\neq L[f] \cdot L[g]$$ You should use :

$$L[t^{n}f(t)] = (-1)^{n}F^{(n)}(s)$$

Where $F(s) = L[f]$ and $F^{(n)} (s)$ is the nth derivative of $F$.

In your case $n=2$

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    I found a formula similar to the one you mentioned GaussTheBauss which I can somewhat understand. Thanks for your help. – 2017-02-23
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$L\{tf(t)\} = \int_0^{\infty} tf(t) e^{-st} dt$

if we integrate both sides with respect to $s.$

$\int L\{tf(t)\} ds = -\int_0^{\infty} f(t) e^{-st} dt = -L\{f(t)\}$

And then we can differentiate both sides (with respect to $s$).

$L\{tf(t)\} = -\frac {d}{ds} L\{f(t)\}$

$L\{t^2\sin t\}=\frac{d^2}{ds^2} \frac {1}{s^2+1}$

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Hint:

$$\mathcal{L}(f(t))=\int_{0}^{\infty} f(t) e^{-st} \ dt.$$

Differentiate under the integral sign twice with respect to $s$ , and we get

$$\frac{\partial^2 \mathcal{L}(f(t))}{\partial s^2}=\int_{0}^{\infty} t^2 f(t) e^{-st} \ dt=\mathcal{L}(t^2 f(t)).$$

What happens when $f(t)=\sin(t)$ ?