 # Need to evaluate the Laplace integral transform (definition below) of sqrt(t) using complex analysis. How is that possible? hat(f)(s)=int_0^(oo) sqrt(t) e^(-st) dt profesorluissp 2022-09-12 Answered
Need to evaluate the Laplace integral transform (definition below) of $\sqrt{t}$ using complex analysis. How is that possible?
$\stackrel{^}{f}\left(s\right)={\int }_{0}^{\mathrm{\infty }}\sqrt{t}{e}^{-st}dt$
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For $s>0$
${\int }_{0}^{\mathrm{\infty }}{t}^{1/2}{e}^{-st}dt={\int }_{0}^{\mathrm{\infty }}\left(u/s{\right)}^{1/2}{e}^{-s\left(u/s\right)}d\left(u/s\right)={s}^{-3/2}{\int }_{0}^{\mathrm{\infty }}{u}^{1/2}{e}^{-u}du\phantom{\rule{0ex}{0ex}}={s}^{-3/2}\mathrm{\Gamma }\left(3/2\right)={s}^{-3/2}\mathrm{\Gamma }\left(1/2\right)1/2={s}^{-3/2}{\pi }^{1/2}/2$
(the last step is with the reflection formula)
Since
${\int }_{0}^{\mathrm{\infty }}{t}^{1/2}{e}^{-st}dt-{s}^{-3/2}{\pi }^{1/2}/2$
is complex analytic for $\mathrm{\Re }\left(s\right)>0$ and it vanishes for $s>0$, it vanishes for $\mathrm{\Re }\left(s\right)>0$ and you got your Laplace transform.

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