 # Find inverse Laplace Transform of : (1)/((s^2+a^2)^2) cochetezgh 2022-09-12 Answered
Find inverse Laplace Transform of : $\frac{1}{\left({s}^{2}+{a}^{2}{\right)}^{2}}$
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As
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
we have
$\frac{d}{ds}F\left(s\right)=\frac{d}{ds}{\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt=-{\int }_{0}^{\mathrm{\infty }}{e}^{-st}tf\left(t\right)dt$
then knowing that ${\mathcal{L}}^{-1}\left(\frac{1}{{s}^{2}+{a}^{2}}\right)=\frac{1}{a}\mathrm{sin}\left(at\right)$
$-\frac{d}{ds}\left(\frac{1}{{s}^{2}+{a}^{2}}\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\frac{t}{a}\mathrm{sin}\left(at\right)dt$
then
$\frac{2s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}⇔\frac{t}{a}\mathrm{sin}\left(at\right)$
and
$\frac{1}{2s}\frac{2s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}⇔\frac{1}{2}{\int }_{0}^{t}\frac{\tau }{a}\mathrm{sin}\left(a\tau \right)d\tau =\frac{\mathrm{sin}\left(at\right)-at\mathrm{cos}\left(at\right)}{2{a}^{3}}$

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$\frac{1}{\left({s}^{2}+{a}^{2}{\right)}^{2}}=\frac{1}{\left(s+ia{\right)}^{2}\left(s-ia{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=-\frac{1}{4{a}^{2}}{\left[\frac{1}{s+ia}-\frac{1}{s-ia}\right]}^{2}\phantom{\rule{0ex}{0ex}}=-\frac{1}{4{a}^{2}}\left[\frac{1}{\left(s+ia{\right)}^{2}}-2\frac{1}{\left(s+ia\right)\left(s-ia\right)}+\frac{1}{\left(s-ia{\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{4{a}^{2}}\left[\frac{1}{\left(s+ia{\right)}^{2}}-\frac{1}{ia}\left(\frac{1}{s-ia}-\frac{1}{s+ia}\right)+\frac{1}{\left(s-ia{\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4{a}^{2}}\frac{d}{ds}\left[\frac{1}{s+ia}+\frac{1}{s-ia}\right]+\frac{1}{4i{a}^{3}}\left(\frac{1}{s-ia}-\frac{1}{s+ia}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{4{a}^{2}}\frac{d}{ds}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+ia\right)t}+{e}^{-\left(s-ia\right)t}dt\phantom{\rule{0ex}{0ex}}+\frac{1}{4i{a}^{3}}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-ia\right)t}-{e}^{-\left(s+ia\right)t}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4{a}^{2}}{\int }_{0}^{\mathrm{\infty }}-t\left({e}^{-\left(s+ia\right)t}+{e}^{-\left(s-ia\right)t}\right)dt\phantom{\rule{0ex}{0ex}}+\frac{1}{4i{a}^{3}}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-ia\right)t}-{e}^{-\left(s+ia\right)t}dt\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\left[\frac{1}{4{a}^{2}}\left(-t{e}^{-iat}-t{e}^{iat}\right)+\frac{1}{4i{a}^{3}}\left({e}^{iat}-{e}^{-iat}\right)\right]dt\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\left[-\frac{1}{2{a}^{2}}t\mathrm{cos}\left(at\right)+\frac{1}{2{a}^{3}}\mathrm{sin}\left(at\right)\right]dt$
So the inverse Laplace transform is
$-\frac{1}{2{a}^{2}}t\mathrm{cos}\left(at\right)+\frac{1}{2{a}^{3}}\mathrm{sin}\left(at\right).$

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