hemotropS7A

2022-11-25

How to find inverse Laplace transform of the following function?
$X\left(s\right)=\frac{s}{{s}^{4}+1}$
I tried to use the definition: $f\left(t\right)={\mathcal{L}}^{-1}\left\{F\left(s\right)\right\}=\frac{1}{2\pi i}\underset{T\to \mathrm{\infty }}{lim}{\int }_{\gamma -iT}^{\gamma +iT}{e}^{st}F\left(s\right)\phantom{\rule{thinmathspace}{0ex}}ds$or the partial fraction expansion but I have not achieved results.

agyaly1F

Expert

Hint: With
${s}^{4}+1={s}^{4}+1+2{s}^{2}-2{s}^{2}=\left({s}^{2}+1-\sqrt{2}s\right)\left({s}^{2}+1+\sqrt{2}s\right)$
then
$\frac{s}{{s}^{4}+1}=\frac{\sqrt{2}}{4}\frac{1}{{s}^{2}+1-\sqrt{2}s}-\frac{\sqrt{2}}{4}\frac{1}{{s}^{2}+1-\sqrt{2}s}$
then complete the squares in denominators!

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