# Does this Laplace transform have a closed form?

I want to know if the Laplace transform of

$$x^\alpha (1+ax)^\beta$$

has any closed form?

I really appreciate your help.

-
Yes, but are you alright with the Kummer confluent hypergeometric function as a "closed form"? – J. M. Aug 7 '11 at 14:29

Yes it does, but requires special functions known as Tricomi confluent hypergeometric function (see functions.wolfram.com). The integral reduces to the one stated in the linked page after change of variables $x = \frac{y}{a}$:
$$LT_s(x^\alpha (1+a x)^\beta) = a^{\alpha - 1} \Gamma(1 + \alpha) U(1 + \alpha, 2 + \alpha + \beta, \frac{s}{a})$$
@SMH: Note that for $\alpha$ and $\beta$ positive integers, the confluent hypergeometric function degenerates to a rational function. – J. M. Aug 7 '11 at 14:41
@SMH: In fact, when $\beta$ is a non-negative integer, the answer is a generalized rational function in $s$. Use binomial theorem for $(1+a x)^\beta$ and integrate term-wise. – Sasha Aug 7 '11 at 16:06
Of course you want to assume $\alpha > -1$ and either $a \ge 0$ or $\beta < 1$ for the Laplace transform to exist. It could also be expressed, according to Maple, as $\Gamma \left( \alpha+1 \right) {{\rm e}^{1/2\,{\frac {s}{a}}}} {{\rm \bf W}\left(1/2\,\beta-1/2\,\alpha,\,1/2+1/2\,\alpha+1/2\,\beta,\,{\frac {s}{a}}\right)} {a}^{1/2\,\beta-1/2\,\alpha}{s}^{-1/2\,\alpha-1/2\,\beta-1}$, where $\rm \bf W$ is a Whittaker W function. – Robert Israel Aug 7 '11 at 23:59