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I'm looking for a solution to $$\int x^{-a} \Gamma\left( b, c x^{-d} \right) dx.$$

Mathematica gives me the following solution, but I'd like to know/understand the steps involved in finding it.

$$\int x^{-a} \Gamma\left( b, c x^{-d} \right) dx = \frac{x^{1-a} \left(\left(c x^{-d}\right)^{\frac{1-a}{d}} \Gamma \left(\frac{a+b d-1}{d},c x^{-d}\right)-\Gamma \left(b,c x^{-d}\right)\right)}{a-1}.$$

The following solution, to a similar integral, seems to be a starting point. However, I have no idea how to proceed from that.

https://functions.wolfram.com/GammaBetaErf/Gamma2/21/01/02/01/

Does someone know how to find this solution?

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    $\begingroup$ just change variables to $z=cx^{-d}$ and then use the integral you link to. $\endgroup$ Feb 7, 2022 at 17:49

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To remove this question from the "unanswered" list; given the indefinite integral $$\int z^{\alpha-1}\Gamma(b,z)\,\mathrm{d}z=\alpha^{-1}[ z^\alpha \Gamma(b, z)-\Gamma(b + \alpha, z)]$$ change variables to $z=cx^{-d}$, $\mathrm{d}z=-dcx^{-d-1}\mathrm{d}x$, hence $$-\int dc^\alpha x^{-\alpha d-1}\Gamma(b,cx^{-d})\,\mathrm{d}x=\alpha^{-1}[ c^\alpha x^{-\alpha d} \Gamma(b, cx^{-d})-\Gamma(b + \alpha, cx^{-d})].$$ Now identify $a=\alpha d+1$ to arrive at $$\int x^{-a}\Gamma(b,cx^{-d})\,\mathrm{d}x=-\frac{1}{a-1}[x^{1-a}\Gamma(b,cx^{-d})-c^{(1-a)/d}\Gamma(b+(a-1)/d,cx^{-d})],$$ which is the desired answer.

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