How do you evaluate the inverse transform below using convolution ? ccL^(−1)[(s)/((s^2+a^2)^2)]

How do you evaluate the inverse transform below using convolution ? ${\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right]$
I tried
$\begin{array}{rl}{\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right]\left(t\right)& =\underset{0}{\overset{t}{\int }}\mathrm{sin}t\cdot \mathrm{cos}\left(at-a\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau =\mathrm{sin}t\cdot \underset{0}{\overset{t}{\int }}\mathrm{cos}\left(at-a\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau \\ & =\mathrm{sin}t\cdot {\left[-\frac{1}{a}\mathrm{sin}\left(at-a\tau \right)\right]}_{0}^{t}=\mathrm{sin}t\cdot \left[0-\left(-\frac{1}{a}\mathrm{sin}\left(at\right)\right)\right]\\ & =\frac{1}{a}{\mathrm{sin}}^{2}\left(at\right)\end{array}$
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siriceboynu1
${\mathcal{L}}^{-1}\left(\frac{{s}^{2}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right)\left(t\right)={\int }_{0}^{t}\mathrm{sin}t\cdot \mathrm{cos}\left(at-a\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau$
you should have
${\mathcal{L}}^{-1}\left(\frac{{s}^{2}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right)\left(t\right)=\frac{1}{a}{\int }_{0}^{t}\mathrm{sin}\left(a\tau \right)\cdot \mathrm{cos}\left(at-a\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau$
because the convolution of functions f,g is defined as
$\left(f\ast g\right)\left(t\right)={\int }_{\mathbb{R}}g\left(\tau \right)\cdot f\left(t-\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau$
where $f\left(t\right):=\mathrm{cos}\left(a\cdot t\right)\cdot {1}_{\left[0,\mathrm{\infty }\right)}\left(t\right)$, $g\left(t\right):=\frac{1}{a}\cdot \mathrm{sin}\left(a\cdot t\right)\cdot {1}_{\left[0,\mathrm{\infty }\right)}\left(t\right)$
Did you like this example?
Keshawn Moran
Assume someone tells you to derivative $\frac{1}{\left({s}^{2}+{a}^{2}\right)}$ respect to s wherein a is a constant. You certainly will reply
$\frac{-2s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}$
${\left(\frac{1}{\left({s}^{2}+{a}^{2}\right)}\right)}^{\prime }=\frac{-2s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}$
or
${\left(\frac{-1}{2\left({s}^{2}+{a}^{2}\right)}\right)}^{\prime }=\frac{s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}$
or
$\frac{1}{2}×\left(-1{\right)}^{1}{\left(\frac{1}{\left({s}^{2}+{a}^{2}\right)}\right)}^{\prime }=\frac{s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}$
But surely know that
$\mathcal{L}\left({t}^{1}f\left(t\right)\right)=\left(-1{\right)}^{1}{F}^{\prime }\left(s\right)$
Now can you find the proper f(t) form tow last equalities when you know that
$\mathcal{L}\left(\frac{1}{a}\mathrm{sin}\left(at\right)\right)=\frac{1}{{s}^{2}+{a}^{2}}$
This another approach besides to yours and first answer.