As $L(e^{at})=\frac1{s-a}$
So putting $a=0,L(1)=\frac1s$
and putting $a=c+id,L(e^{(c+id)t})=\frac1{s-(c+id)}$
so, $L(e^{ct}\cos dt)+iL(e^{ct}\sin dt)=\frac{s-c+id}{(s-c)^2+d^2}$
$cos^2t=\frac{1+\cos2t}2$
So, $L(cos^2t)=\frac12 L(1)+\frac12 L(\cos2t)=\frac1{2s}+\frac s{2(s^2+2^2)}$
$\cos3t\sin3t=\frac{\sin6t}2$
So, $L(\cos3t\sin3t)=\frac{L(\sin6t)}2=\frac{6}{2(s^2+6^2)}$
We know, $L(e^{at}f(t))=F(s-a)$ where $F(s)=L(f(t))$
So, $L(e^{at} t^n)=\frac{n!}{(s-a)^{n+1}}$
Hence $L(te^t)=\frac1{(s-a)^2}$ putting $a=n=1$
Putting $a=2i,n=1; L(e^{2it} t)=\frac1{(s-2i)^2}$
$L(t\cos 2t)+iL(t\sin 2t)=\frac{(s+2i)^2}{(s^2+4)^2}=\frac{s^2-4+i4s}{(s^2+4)^2}$
Now compare the real and the imaginary parts.
We know that
$$ \displaystyle \mathcal{L} \{t^{n-1}\} = \frac{\Gamma(n)}{s^{n}}, n>0 ,s>0 $$
$$ \frac{1}{s^n} = \frac{\displaystyle \mathcal{L} \{t^{n-1}\}}{\Gamma(n)} $$
$$ \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^n}\} = \frac{t^{n-1}}{\Gamma(n)} $$
Therefore
$$ \displaystyle \mathcal{L^{-1}} \{ s^{-1/2} e^{-1/s} \} = \displaystyle \mathcal{L^{-1}} \{ \frac{1}{s^{1/2}} - \frac{1}{s^{3/2}} + \frac{1}{(2!s^{5/2})} + ... + (-1)^{n}\frac{1}{n! s^{n+\frac{1}{2}}} \} $$
$$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^{n+\frac{1}{2}}}\} $$
$$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{t^{n-\frac{1}{2}}}{\Gamma(n+\frac{1}{2})} $$
But after this I don't know how to simplify
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}\expo{-\root{s}}\expo{ts} \,{\dd s \over 2\pi\ic} = -\int_{-\infty}^{0}\expo{-\root{-s}\exp\pars{\pi\ic/2}}\expo{ts} \,{\dd s \over 2\pi\ic} - \int_{0}^{-\infty}\expo{-\root{-s}\exp\pars{-\pi\ic/2}}\expo{ts} \,{\dd s \over 2\pi\ic} \\[5mm] = &\ -\int_{0}^{\infty}\expo{-\root{s}\ic}\expo{-ts} \,{\dd s \over 2\pi\ic} + \int_{0}^{\infty}\expo{\root{s}\ic}\expo{-ts} \,{\dd s \over 2\pi\ic} = {1 \over \pi}\int_{0}^{\infty}\sin\pars{\root{s}}\exp\pars{-ts}\,\dd s \\[5mm] = &\ {2 \over \pi}\int_{0}^{\infty}\sin\pars{s}\exp\pars{-ts^{2}}s\,\dd s = \bbx{\ds{{1 \over 2\root{\pi}}\,{\exp\pars{-1/\bracks{4t}} \over t^{3/2}}}} \end{align}
\begin{align} &{2 \over \pi}\int_{0}^{\infty}\sin\pars{s}\exp\pars{-ts^{2}}s\,\dd s = \left.-\,{2 \over \pi}\,\partiald{}{\mu}\int_{0}^{\infty} \cos\pars{\mu s}\exp\pars{-ts^{2}}\,\dd s\,\right\vert_{\ \mu\ =\ 1} \\[5mm] = &\ \left.-\,{1 \over \pi}\,\partiald{}{\mu}\Re\int_{-\infty}^{\infty} \exp\pars{-ts^{2} + \mu s\ic}\,\dd s\,\right\vert_{\ \mu\ =\ 1} = -\,{1 \over \pi}\,\partiald{}{\mu} \bracks{\root{\pi \over t}\exp\pars{-\,{\mu^{2} \over 4t}}}_{\ \mu\ =\ 1} \end{align}