Question Number 219236 by Nicholas666 last updated on 20/Apr/25
$$ \\ $$$$\:\:\:\:{f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\int_{\:{c}−{i}\infty} ^{\:{c}+{i}\infty} \:\frac{{e}^{{st}} }{{s}^{{k}} \:}\:\:{ds}\:\:\:,\:\:{k}\:\in\mathbb{C} \\ $$$$\: \\ $$
Commented by Nicholas666 last updated on 22/Apr/25
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Answered by SdC355 last updated on 21/Apr/25
$$\mathcal{L}^{−\mathrm{1}} =\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}{e}^{{st}} \:,\:{t}\in\mathbb{R} \\ $$$$\mathcal{L}_{{t}} ^{−\mathrm{1}} \left\{\frac{{n}!}{{s}^{{n}+\mathrm{1}} }\right\}={t}^{{n}} \:\:\therefore\:\:\:\mathcal{L}_{{t}} ^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{s}^{{k}} }\right\}=\frac{{t}^{{k}} }{\boldsymbol{\Gamma}\left({k}\right)} \\ $$
Commented by Nicholas666 last updated on 22/Apr/25
$${thanks} \\ $$