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6.51 Example continued

None of the standard Laplace transforms on frame 6.44 equals $s/(s^{2}+\alpha^{2})^{2}$ . To complete the problem, we use the following result.

Theorem.

Suppose the Laplace transform of $f(x)$ converges for all $s>s_{0}$ and equals $F(s)$ . Then the Laplace transform of $xf(x)$ exists for all $s>s_{0}$ and ${\mathcal{L}}(xf(x))=-\frac{dF}{ds}$ . Hence ${\mathcal{L}}^{-1}(\frac{dF}{ds})=-x{\mathcal{L}}(F)$ .

Now

\frac{s}{(s^{2}+\alpha^{2})^{2}}={-\frac{1}{2}\frac{d}{ds}\left(\frac{1}{s^{2}% +\alpha^{2}}\right).}

Using the Theorem, we see that the solution to the problem is the inverse Laplace transform of $\frac{s}{(s^{2}+\alpha^{2})^{2}}$ , which is

{\frac{1}{2}x{\mathcal{L}}^{-1}\left(\frac{1}{s^{2}+\alpha^{2}}\right)=}\,{% \frac{1}{2\alpha}x\sin\alpha x.}