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$f(t) = a + b \exp(-c \cdot t ^ d) $, where $a,b,c,d$ are constants, and $d$ is power of $t$.

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up vote 2 down vote accepted

It boils down to computing $J = \int_0^\infty \exp(-st - t^d)\, dt$. I'll assume $d > 1$. Write as a series in $-s$: $J = \sum_{n=0}^\infty \int_0^\infty \frac{(-s)^n t^n}{n!} \exp(-t^d)\, dt = \sum_{n=0}^\infty \frac{\Gamma((n+1)/d)}{n!\, d} (-s)^n$. I don't know if there is a "closed form" for this, but for each positive integer $d$ it can be written in terms of hypergeometric functions.

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Apparently, as long as $d$ is rational and bigger than 1, the expression is hypergeometric. The expressions are rather unwieldy here that a numerical method might make more sense for computational needs. – J. M. May 16 '11 at 5:04
Thanks for the answer given by @Robert Israel and @J.M.. But the answer of the question is $(a+bexp(−c*t^d))/s$ computed by the mathematic tool of Maple. As you find, the answer contains $t$. So, what is the right answer of the problem? – zghu001 May 16 '11 at 11:11
@zghu: What, *exactly*, did you input into Maple? – J. M. May 16 '11 at 11:22
@J.M.. I have put $a+bexp(-ct^d)$ into Maple, and got that result. – zghu001 May 16 '11 at 14:54
@J.M.. I know what has happend. In $a+bexp(-ct^d)$, I missed the multiplication between $b$ and $exp(-ct^d)$, also the multiplication between $c$ and $t^d$. Thus, the wrong answer was gotted. Thank you very much! So, the question doesn't have answer? – zghu001 May 16 '11 at 15:25

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