I'm looking for a solution or approximation to the following indefinite integral $$\int x^{-a} \Gamma\left( b, c x^{-d} + e \right) dx.$$
I've tried Mathematica, but it does not converge to a solution. Is there a way to solve it?
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- 1$\begingroup$ there is a closed-form solution for $e=0$, but not for nonzero $e$ (and arbitrary exponents, $a,d$) $\endgroup$Carlo Beenakker– Carlo Beenakker2023-06-16 11:51:41 +00:00Commented Jun 16, 2023 at 11:51
- 1$\begingroup$ The mentioned closed form solution for $e=0$ is here mathoverflow.net/questions/415584/… $\endgroup$Felipe Augusto de Figueiredo– Felipe Augusto de Figueiredo2023-06-16 11:57:14 +00:00Commented Jun 16, 2023 at 11:57
- 1$\begingroup$ @CarloBeenakker, would an approximation be possible? $\endgroup$Felipe Augusto de Figueiredo– Felipe Augusto de Figueiredo2023-06-16 11:58:07 +00:00Commented Jun 16, 2023 at 11:58
- 1$\begingroup$ For $x\to\infty$, use Taylor series in $z$ of $\Gamma(b, z+e)$ and integrate term by term. For $y=(c x^{-d}+e) \to \infty$, use asymptotic expansion of $\Gamma(b,y)$ and integrate term by term. Or do you need something more sophisticated? $\endgroup$Igor Khavkine– Igor Khavkine2023-06-17 13:08:25 +00:00Commented Jun 17, 2023 at 13:08
- 2$\begingroup$ Your question is missing information about what regime of approximation you are interested in. Away from singular limits, where some parameter is going to infinity, you can also just apply numerical integration. $\endgroup$Igor Khavkine– Igor Khavkine2023-06-18 00:21:52 +00:00Commented Jun 18, 2023 at 0:21
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1 Answer
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4 *Could this be workable ? *
$$I=\int x^{-a} \Gamma\left( b, c x^{-d} + e \right) \,dx$$ Let $c x^{-d}=t$ and $\alpha=\frac{a-d-1}{d}$ to make $$I=-\frac 1 d \,c^{\frac{1-a}{d}}\int t^\alpha \,\,\Gamma (b,t+e)\,dt$$ Expand as a series of $e$ around $e=0$ $$\Gamma (b,t+e)=\sum_{n=0}^\infty A_n\, e^n$$ where
$$A_n=-\frac{(n-1) (t+n-b-1)\,A_{n-1}+(n-2)\,A_{n-2}}{n(n-1)t }$$ with $$A_0=\Gamma (b,t) \qquad \qquad A_1=-e^{-t}\, t^{b-1}$$
- $\begingroup$ Please, complete the answer. $\endgroup$Felipe Augusto de Figueiredo– Felipe Augusto de Figueiredo2024-01-22 10:38:16 +00:00Commented Jan 22, 2024 at 10:38
- $\begingroup$ @FelipeAugustodeFigueiredo? My very first line is a question to you. Did you try something ? I am curious to know $\endgroup$Claude Leibovici– Claude Leibovici2024-01-22 10:44:26 +00:00Commented Jan 22, 2024 at 10:44
- $\begingroup$ Dear, I haven't tried anything yet. However your answer seems to be the way to a solution. That is why I asked if you could go on with your answer. Thanks. $\endgroup$Felipe Augusto de Figueiredo– Felipe Augusto de Figueiredo2024-01-23 11:24:44 +00:00Commented Jan 23, 2024 at 11:24
- $\begingroup$ @FelipeAugustodeFigueiredo. Please try it and let us discuss later $\endgroup$Claude Leibovici– Claude Leibovici2024-01-23 12:51:42 +00:00Commented Jan 23, 2024 at 12:51