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I'm working in asymptotic expansions and it appear this integral

$$\int_0^{\infty}e^{-t^{2m}}t^n(t-a)^{s}dt,\quad a\in\mathbb{R}$$

where $m,n\in\mathbb{N}$ and $s\in(-1,0]$.

When $a=0$, it becomes a Gamma function. I wonder if it can be written in terms of some special function when $a\neq0$.

Any help will be welcomed.

amWhy
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popi
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    Use \quad or \qquad for spacing, instead of so many \,s. FYI, \, is a thin space, \ is a space, \; is a tad thicker of a space. But for sizeable spacing, use \quad or \qquad. Try it in practice. (I used \quad to edit your post above.) – amWhy Jan 15 '21 at 23:11
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    There is a closed form in terms of the Meijer G-function. But if the question is about asymptotics, finding a closed form may be unnecessary. – Maxim Jan 23 '21 at 22:10
  • Dear Maxim, can you write or say something about this closed form in terms of the Meijer G-function? – popi Jan 24 '21 at 10:58
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    It's this method. The Mellin transforms are $$F(p) = \mathcal Mt \mapsto e^{-t^{2 m}} = \frac 1 {2 m} \Gamma {\left( \frac p {2 m} \right)}, \ G(p) = \mathcal Mt \mapsto (t + 1)^s = \frac 1 {\Gamma(-s)} \Gamma(p) \Gamma(-p - s),$$ so for $a < 0$ we get $$\mathcal M[t \mapsto e^{-t^{2 m}} (t - a)^s](n + 1) = (-a)^s \mathcal M^{-1}[p \mapsto F(n + 1 - p) G(p)] {\left( -\frac 1 a \right)}.$$ The Fox H-function can be converted to a G-function by applying the multiplication theorem to $\Gamma(2 m (p + p_0))$. – Maxim Jan 24 '21 at 17:17

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