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In question 1) we get Laplace transform of $$ g(t) = t^a $$ is:

$$ \hat g(t)= {1/s^{a+1}}\int_0^\infty e^{-t}x^a $$

then I was stuck at question 2) which asks me to evaluate the inverse laplace transform of $ \hat g(p) $ which is

$$ {1/2\pi i}\int_0^\infty e^{pt}\hat g(p)dp $$

I know the answer should be $ t^a $ as the inverse transform comes back to itself, but I cannot figure out how to make the contour integration. I tried to apply Cauchy's residue theorem to eliminate the $ 1/2 \pi i $ but was stuck then. Thanks a lot for help!

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I don't understand the form of $\hat{g}(t)$... – mwoua May 3 '13 at 20:54

1 Answer 1

up vote 2 down vote accepted

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm g}\pars{t}\equiv t^{a}\quad\imp\quad \hat{\rm g}\pars{s} = \int_{0}^{\infty}t^{a}\expo{-st}\,\dd t ={\Gamma\pars{a + 1} \over s^{a + 1}}}$ where $\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$. Also, $\ds{{\rm g}\pars{t} =\int_{\gamma - \infty\ic}^{\gamma + \infty\ic} {\Gamma\pars{a + 1} \over s^{a + 1}}\,\expo{st}\,{\dd s \over 2\pi\ic}\,, \qquad\gamma > 0}$.

\begin{align} {\rm g}\pars{t}& =\Gamma\pars{a + 1}\int_{\gamma - \infty\ic}^{\gamma + \infty\ic} {\expo{st} \over s^{a + 1}}\,{\dd s \over 2\pi\ic}=\Gamma\pars{a + 1}\times \\[3mm]&\bracks{% -\int_{-\infty}^{0}\pars{-s}^{-a - 1}\expo{-\pars{a + 1}\pi\ic}\expo{st} {\dd s \over 2\pi\ic} -\int_{0}^{-\infty}\pars{-s}^{-a - 1}\expo{\pars{a + 1}\pi\ic}\expo{st} {\dd s \over 2\pi\ic}} \\[3mm]&=\Gamma\pars{a + 1}\bracks{% \expo{-\pi a\ic}\int_{0}^{\infty}s^{-a - 1}\expo{-st} {\dd s \over 2\pi\ic} -\expo{\pi a\ic}\int_{0}^{\infty}s^{-a - 1}\expo{-st}{\dd s \over 2\pi\ic}} \\[3mm]&=-\,{1 \over \pi}\,\Gamma\pars{a + 1}\, {\expo{\pi a\ic} - \expo{-\pi a\ic} \over 2\ic} \int_{0}^{\infty}s^{-a - 1}\expo{-st}\dd s \\[3mm]&=-\,{\Gamma\pars{a + 1} \over \pi}\,\sin\pars{\pi a}t^{a}\ \underbrace{\int_{0}^{\infty}s^{-a - 1}\expo{-s}\dd s}_{\ds{=\ \Gamma\pars{-a}}} \end{align}

$$ {\rm g}\pars{t} ={\Gamma\pars{1 + a}\Gamma\pars{-a}\sin\pars{-\pi a} \over \pi}\,t^{a} $$

With Euler Reflection Formula ${\bf\mbox{6.1.17}}$, $\ds{{\Gamma\pars{1 + a}\Gamma\pars{-a}\sin\pars{-\pi a} \over \pi} = 1}$ such that

$$ \color{#44f}{\large{\rm g}\pars{t} = t^{a}} $$

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