The, kind of, truncated moment of the generalized Gamma distribution for positive $a,d$ and $p$ is $$\int_0^\infty (x-k)_+^a x^{d-1}e^{-x^p}dx=\frac1p\int_0^\infty (y^{\frac1p}-k)_+^a\,y^{\frac dp-1}e^{-y}\,dy. \tag1$$ The truncated moment of the Gamma distribution can be transformed into the confluent hypergeometric function of the second kind defined as such. Is there a similar special function transform for integral $(1)$?
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There is a closed form in terms of the Fox H-function, $$\int_k^\infty x^{d - 1} (x - k)^a e^{-x^p} dx = k^{a + d} \hspace {1.5px} \Gamma(a + 1) H_{1, 2}^{2, 0} {\left( k^p \middle| {(1 - d, p) \atop (0, 1), (-a - d, p)} \right)}, \ k > 0, ; p > 0, ; \operatorname {Re} a > -1.$$ – Maxim Jun 04 '20 at 05:00
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@Maxim: Could you please give a detailed derivation as an answer? – Hans Jun 04 '20 at 06:06
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1It's an application of this method. We need $$\mathcal Mx \mapsto e^{-x} = \Gamma(s), \ \mathcal Mx \mapsto (x - k)_+^a = \frac {k^{s + a} \Gamma(a + 1) \Gamma(-a - s)} {\Gamma(1 - s)}.$$ – Maxim Jun 04 '20 at 17:49