Inverse laplace with fractional order

Question

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$$ L^{-1}\left[ \frac{s^{\alpha}}{s^2\alpha+a^2}\right]\\ L^{-1}\left[ \frac{a}{s^2\alpha+a^2}\right] $$

Can any one help me with those where alpha is fraction/rational number ... thanks in advance

Answer 1

From this PDF on page 26:

$$\mathcal{L}_s^{-1}\left[\frac{s^{\alpha }}{s^{\beta }+a^2}\right](t)=t^{-1-\alpha +\beta } E_{\beta ,-\alpha +\beta }\left(-a^2 t^{\beta }\right)$$

$$\mathcal{L}_s^{-1}\left[\frac{a}{s^{\beta }+a^2}\right](t)=a t^{-1+\beta } E_{\beta ,\beta }\left(a^2 t^{\beta }\right)$$

where: $E_{\beta ,\beta -\alpha }\left(-a^2 t^{\beta }\right)$ and $E_{\beta ,\beta }\left(a^2 t^{\beta }\right)$ is MittagLefflerE function.