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I want to know if the Laplace transform of

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

has any closed form?

I really appreciate your help.

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

up vote 2 down vote accepted

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 $x = \frac{y}{a}$:

$$ LT_s(x^\alpha (1+a x)^\beta) = a^{\alpha - 1} \Gamma(1 + \alpha) U(1 + \alpha, 2 + \alpha + \beta, \frac{s}{a}) $$

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@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
    
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