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Est-ce qu'il y a une fonction spéciale sous la forme suivante: $$\int e^{-st}e^{-(\alpha/(1+t))}t^{\beta-1}(1+t)^{-(\beta-1)-\gamma}dt$$

ou $$\int e^{-st}e^{-(\alpha/t)}t^{\beta-1}(1-t)^{\gamma-2}dt$$

[Edit added by A. Rekalo: Is there a special function of the form $\int...$ or $\int...$?]

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Could you phrase your question in english please? –  J.C. Ottem Feb 10 '11 at 16:15
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"Is there a special function which has the following form (first formula) or the following (second formula). –  Igor Rivin Feb 10 '11 at 16:28
    
I assume the integrals are definite, from zero to infinity. –  Igor Rivin Feb 10 '11 at 16:29
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J. C. Ottem- We already had a long discussion about this (tea.mathoverflow.net/discussion/142/nonenglish-posts) but the consensus at the end was that if people want to write in French, that's up to them, but that it was advisable (as A. Rekalo did) to add an English translation. –  Ben Webster Feb 10 '11 at 17:16
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In this particular case, it seems hard to not get what the OP is asking (the fact that the math is a little funky is a whole'nother matter). –  Igor Rivin Feb 10 '11 at 17:36
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2 Answers

Original proof:

Let's consider the second integral with $\alpha = 0$. Noting, on the one hand, that the integrand is not real for $t \gt 1$ if, for example, $\gamma=2.5$, and, on the other hand, observing its relation to the density function of the ${\rm Beta}(\beta,\gamma-1)$ distribution, it seems likely that the integral (with $\alpha = 0$) is actually $$ I(s;\beta,\gamma) = \int_0^1 {e^{ - st} t^{\beta - 1} (1 - t)^{\gamma - 2} \,{\rm d}t}, $$ where $\beta \gt 0$ and $\gamma \gt 1$. Now, $\frac{{\Gamma (\beta + \gamma - 1)}}{{\Gamma (\beta )\Gamma (\gamma - 1)}} I(s;\beta,\gamma)$, for $s \geq 0$, is the Laplace transform of the ${\rm Beta}(\beta,\gamma-1)$ distribution. The moment-generating function is given by $$ m(s) = 1 + \sum\limits_{k = 1}^\infty {\bigg(\prod\limits_{r = 0}^{k - 1} {\frac{{\beta + r}}{{\beta + \gamma - 1 + r}}} \bigg)\frac{{s^k }}{{k!}}}. $$ Writing $$ \frac{{\Gamma (\beta + \gamma - 1)}}{{\Gamma (\beta )\Gamma (\gamma - 1)}} I(s;\beta,\gamma) = m(-s) $$ gives $$ I(s;\beta,\gamma) = \frac{{\Gamma (\beta )\Gamma (\gamma - 1)}}{{\Gamma (\beta + \gamma - 1)}}\bigg[1 + \sum\limits_{k = 1}^\infty {\bigg(\prod\limits_{r = 0}^{k - 1} {\frac{{\beta + r}}{{\beta + \gamma - 1 + r}}} \bigg)\frac{{(-s)^k }}{{k!}}}\bigg]. $$

It turns out that this result is well known (cf. Eq. (3) here):

Using the standard notation in the theory of special functions (in particular the hypergeometric function), $I(s;\beta,\gamma)$ can be written as $$ I(s;\beta,\gamma) = \frac{{\Gamma (\beta )\Gamma (\gamma - 1)}}{{\Gamma (\beta + \gamma - 1)}} \bigg[ 1 + \sum\limits_{k = 1}^\infty {\frac{{(\beta )_k }}{{(\beta + \gamma - 1)_k }}\frac{{( - s)^k }}{{k!}}} \bigg] $$ ($(x)_n$ represents the rising factorial). Finally, in terms of the confluent hypergeometric function of the first kind, $$ I(s;\beta,\gamma) = \frac{{\Gamma (\beta )\Gamma (\gamma - 1)}}{{\Gamma (\beta + \gamma - 1)}} {}_1F_1 (\beta ;\beta + \gamma - 1; - s). $$

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The first one, in special case $\alpha=1, \beta=1,\gamma=0$, namely $$ \int_{0}^{\infty} \operatorname{e} ^{-s t} \operatorname{e} ^{-\frac{1}{1 + t}} \;d t , $$ is not recognized by Maple.

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