# How to find the Laplace transformation L{cos(8t) sqrt(t/(pi))}?

Given $\mathcal{L}\left\{\mathrm{cos}\left(8t\right)/\sqrt{\pi t}\right\}={e}^{-8/s}/\sqrt{s}$
How to find the Laplace transformation
$\mathcal{L}\left\{\mathrm{cos}\left(8t\right)\sqrt{t/\pi }\right\}$?
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Aldo Harrington
$\mathcal{L}\left\{\frac{\mathrm{cos}\mathcal{4}\sqrt{\mathcal{2}\mathcal{t}}}{\sqrt{\pi \mathcal{t}}}\mathcal{u}\mathcal{\left(}\mathcal{t}\mathcal{\right)}\right\}\mathcal{=}\frac{{\mathcal{e}}^{\mathcal{-}\mathcal{8}\mathcal{/}\mathcal{s}}}{\sqrt{\mathcal{s}}}\mathcal{.}$
Since
$\frac{\sqrt{t}\mathrm{cos}4\sqrt{2t}}{\sqrt{\pi }}=\frac{\mathrm{cos}4\sqrt{2t}}{\sqrt{\pi t}}\cdot t,$
Integrating-by-parts $⟹$ Laplace transform of $t\cdot f\left(t\right)$ is $-\mathcal{L}\left\{\mathcal{f}\mathcal{\left(}\mathcal{t}\mathcal{\right)}\right\}$
We have:
$\mathcal{L}\left\{\frac{\sqrt{\mathcal{t}}\mathrm{cos}\mathcal{4}\sqrt{\mathcal{2}\mathcal{t}}}{\sqrt{\pi }}\mathcal{u}\mathcal{\left(}\mathcal{t}\mathcal{\right)}\right\}\mathcal{=}\mathcal{-}\frac{\mathcal{d}}{\mathcal{d}\mathcal{s}}\mathcal{L}\left\{\frac{\mathrm{cos}\mathcal{4}\sqrt{\mathcal{2}\mathcal{t}}}{\sqrt{\pi \mathcal{t}}}\mathcal{u}\mathcal{\left(}\mathcal{t}\mathcal{\right)}\right\}\mathcal{=}\mathcal{-}\frac{\mathcal{d}}{\mathcal{d}\mathcal{s}}\frac{{\mathcal{e}}^{\mathcal{-}\mathcal{8}\mathcal{/}\mathcal{s}}}{\sqrt{\mathcal{s}}}\mathcal{=}\frac{{\mathcal{e}}^{\mathcal{-}\mathcal{8}\mathcal{/}\mathcal{s}}\mathcal{\left(}\mathcal{s}\mathcal{-}\mathcal{16}\mathcal{\right)}}{\mathcal{2}{\mathcal{s}}^{\mathcal{5}\mathcal{/}\mathcal{2}}}\mathcal{.}$