Determination of inverse laplace transform using primitive functions

Question

In How can you prove that a function has no closed form integral?, the accepted answer points to http://www.sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf where one can find a corollary by Liouville providing sufficient and necessary conditions for the integral:

$$\int{f(t)e^{g(t)}\mathrm dt}$$

to be determinable (using elementary functions and operations). Now, the inverse Laplace transform of $s^{a-1}/s^a+\lambda$ is given by:

$$\mathcal{L}\left\{ \frac{s^{a-1}}{s^a+\lambda} \right\} = E_{a,1}(-\lambda t)$$

where $E_{a,b}(t)$ is the two-parameter Mittag-Leffler function which is not a primitive function and is defined by:

$$E_{a,b}(t)=\sum_{k=0}^{\infty}{\frac{t^k}{\Gamma(ak+b)}}$$

Can we use Liouville's corollary or some other result to prove that $s^{a-1}/s^a+\lambda$ does not have an inverse Laplace transform that can be expressed in terms of elementary functions and operations? Actually we need to prove that the following function is not an elementary primitive:

$$\varphi(t)=\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT} e^{st} \frac{s^{a-1}}{s^a+\lambda}\mathrm ds$$

Answer 1

The Risch algorithm can determine whether or not a function has an elementary antiderivative. Definite integrals, however, are a completely different matter, and I don't think there is much of a theory about when they can be expressed as elementary functions.

Answer 2

(This should be a comment.)

Note, however, that for special values of $a$ in the Mittag-Leffler function, your series can be expressed in terms of more familiar functions like the exponential, the hyperbolic cosine, or the error function.

On the other hand, there are cases where the indefinite integral is non-elementary, while the definite integral (through happy coincidence or some such) reduces to something completely elementary. Risch won't catch those cases.