# Inverse Laplace Transform by contour integration

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

$${\rm g}\pars{t} ={\Gamma\pars{1 + a}\Gamma\pars{-a}\sin\pars{-\pi a} \over \pi}\,t^{a}$$
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
$$\color{#44f}{\large{\rm g}\pars{t} = t^{a}}$$