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S May 6, 2021 at 14:27 history bounty ended AndreaPaco
S May 6, 2021 at 14:27 history notice removed AndreaPaco
May 4, 2021 at 7:10 comment added AndreaPaco @CarloBeenakker many thanks!
May 4, 2021 at 6:35 comment added Carlo Beenakker @JohannesTrost --- I added the closed-form expressions for $k=2,3,4,5$ --- I hope you can identify the pattern and convert those into the full result!
May 3, 2021 at 22:07 comment added AndreaPaco @PeterTaylor, no, the plots in the domain $x\in(0,1)$ simply do not match. Johannes, if you manage to further simplify the expression found by Carlo please let me know.
May 3, 2021 at 20:46 comment added Johannes Trost Note that the regularized Gaussian Hypergeometric function with $a=-k-1$, a negative integer, is a polynomial in $x$ of order $k+1$, and also its derivative for the third parameter $c$ (even if it is the set to $c=-k$) is. So the first term in the parentheses of Carlo's second equation is a finite sum. When writing the hypergeometric function as its defining sum, the derivative for the first parameter (i.e., the second term in the parentheses of the final formula of Carlo Beenaker) seems to collapse to : $(-1)^{k} x^{k+1} (k+1)! (\ln(1-x)+H_{k+1})$ , but I have to check the details.
May 3, 2021 at 14:14 comment added Carlo Beenakker The $x$-dependence is more complicated: For $x\rightarrow 1$ I find that $x^{-k}I_k(x)\rightarrow-\ln(1-x)$. For $x\rightarrow 0$ I find $x^{-k}I_k(x)\rightarrow -[(k+1)x^{k+1}]^{-1}$.
Apr 29, 2021 at 17:26 vote accept AndreaPaco
Apr 29, 2021 at 12:21 answer added Carlo Beenakker timeline score: 4
S Apr 29, 2021 at 9:17 history bounty started AndreaPaco
S Apr 29, 2021 at 9:17 history notice added AndreaPaco Draw attention
Apr 27, 2021 at 5:49 history edited Francesco Polizzi CC BY-SA 4.0
deleted 11 characters in body; edited title
Apr 26, 2021 at 22:40 history asked AndreaPaco CC BY-SA 4.0