Fractional Calculus - Laplace Transform

Laplace Transform

We can also come at the question via the Laplace transform. Noting that

and

etc., we assert

.

For example

\begin{array}{lcr} J^\alpha\left(t^k\right) &= &\mathcal L^{-1}\left\{\dfrac{\Gamma(k+1)}{s^{\alpha+k+1}}\right\}\\ &= &\dfrac{\Gamma(k+1)}{\Gamma(\alpha+k+1)}t^{\alpha+k} \end{array}

as expected. Indeed, given the convolution rule (and shorthanding for clarity) we find that

\begin{array}{rcl} (J^\alpha f)(t) &= &\frac{1}{\Gamma(\alpha)}\mathcal L^{-1}\left\{\left(\mathcal L\{p\}\right)(\mathcal L\{f\})\right\}\\ &=&\frac{1}{\Gamma(\alpha)}(p*f)\\ &=&\frac{1}{\Gamma(\alpha)}\int_0^t p(t-\tau)f(\tau)\,d\tau\\ &=&\frac{1}{\Gamma(\alpha)}\int_0^t(t-\tau)^{\alpha-1}f(\tau)\,d\tau\\ \end{array}

which is what Cauchy gave us above.

Laplace transforms "work" on relatively few functions, but they are often useful for solving fractional differential equations.

Read more about this topic:  Fractional Calculus

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