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Question Number 130198 by mnjuly1970 last updated on 23/Jan/21

               ....nice  calculus...      calculate::        L [e^(−t) . (√(t )) ] =_(transform) ^(Laplace)  ? ...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}... \\ $$$$\:\:\:\:{calculate}:: \\ $$$$\:\:\:\:\:\:\mathscr{L}\:\left[{e}^{−{t}} .\:\sqrt{{t}\:}\:\right]\:\underset{{transform}} {\overset{{Laplace}} {=}}\:?\:... \\ $$$$\:\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 23/Jan/21

∫_0 ^∞ e^(−(s+1)t) (√t) dt       =(1/((s+1)^(3/2) ))Γ((3/2))=(√(π/(4(s+1)^3 )))

$$\int_{\mathrm{0}} ^{\infty} {e}^{−\left({s}+\mathrm{1}\right){t}} \sqrt{{t}}\:{dt}\:\:\:\:\: \\ $$$$=\frac{\mathrm{1}}{\left({s}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\sqrt{\frac{\pi}{\mathrm{4}\left({s}+\mathrm{1}\right)^{\mathrm{3}} }} \\ $$

Commented by mnjuly1970 last updated on 23/Jan/21

thanks alot..

$${thanks}\:{alot}.. \\ $$

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