Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.