s2vunov

2022-09-04

If F(s) is the Laplace transform of $f\left(t\right)={t}^{-3/2}{\mathrm{e}}^{-1/t}$, show that ${F}^{\prime }\left(s\right)=-{s}^{-1/2}F\left(s\right)$

Quinn Hansen

$\begin{array}{rl}F\left(s\right)& ={\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}{t}^{-3/2}{e}^{-\left(st+1/t\right)}\\ & ={\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}{t}^{-3/2}{e}^{-{s}^{1/2}\left[{s}^{1/2}t+1/\left({s}^{1/2}t\right)\right]}\\ & ={s}^{1/4}{\int }_{0}^{\mathrm{\infty }}du\phantom{\rule{thinmathspace}{0ex}}{u}^{-3/2}{e}^{-{s}^{1/2}\left(u+1/u\right)}\\ & ={s}^{1/4}{\int }_{0}^{\mathrm{\infty }}dv\phantom{\rule{thinmathspace}{0ex}}{v}^{-1/2}{e}^{-{s}^{1/2}\left(v+1/v\right)}\\ & ={s}^{1/2}{\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}{t}^{-1/2}{e}^{-\left(st+1/t\right)}\end{array}$
In the second line, I subbed $t={s}^{-1/2}u$. In the third line, I subbed $u=1/v$. In the fourth line I subbed $v={s}^{1/2}t$. Now we take the derivative of the original integral wrt s:
${F}^{\prime }\left(s\right)=-{\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}{t}^{-1/2}{e}^{-1/t}{e}^{-st}$
Thus,
${F}^{\prime }\left(s\right)=-{s}^{-1/2}F\left(s\right)$

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