I am trying to solve the following integral $$ \int_{0}^\infty x^d \left( \frac{e^{- a x} (b+cx)^{d}}{\left( 1- \frac{\Gamma(d, b+cx)}{\Gamma(d)} \right)}\right)^s dx $$ where $a>0$, $b>0$, $c>0$, $d \in \mathbb{Z}^+$, $s\in \mathbb{R}$ and $\Gamma(\cdot, \cdot)$ denotes the upper-incomplete Gamma function.

I have noticed that the Gamma ratio is negligible for $c$ and $b$ not too small and then by discarding that ratio, I am able to solve the integral in semi-analytical form which include hypergeometric functions.

I would of course rather have an exact expression but I am not sure if it's possible to obtain. If anyone have any ideas on how to go about, I would really like to know about it.


1 Answer 1


Be $\,a,b,c,d,s>0\,$ , $\,d\,$ an integer.

$(A)\hspace{1cm}$ Only an approximation for large positive $\,s\,$ .

Be $\,cd<ab\,$ . Be $\,s\,$ large enough so that $\,\displaystyle \left(1+\frac{x}{s}\right)^s\approx e^x$ and $\,\displaystyle \Gamma\left(x,y+\frac{z}{s}\right)\approx \Gamma(x,y)\,$ .

It's $\,\displaystyle \left(1-\frac{\Gamma(d,b+\frac{x}{n})}{\Gamma(d)}\right)^n\to 0\,$ for $\,n\to\infty\,$ .

Then we have

$\displaystyle \int\limits_0^\infty x^d (b+cx)^{ds} \left(1-\frac{\Gamma(d,b+cx)}{\Gamma(d)}\right)^{-s} e^{-asx}dx = $

$\displaystyle =\frac{b^{ds}}{(as)^{d+1}} \int\limits_0^\infty x^d \left(1+\frac{c}{abs}x\right)^{ds} \left(1-\frac{\Gamma(d,b+\frac{c}{as}x)}{\Gamma(d)}\right)^{-s} e^{-x}dx$

$\displaystyle \approx \frac{b^{ds}}{(as)^{d+1}\left(1-\frac{\Gamma(d,b)}{\Gamma(d)}\right)^s} \int\limits_0^\infty x^d e^{-(1-\frac{cd}{ab})x}dx = \frac{b^{ds}d!}{\left(as(1-\frac{cd}{ab})\right)^{d+1}\left(1-\frac{\Gamma(d,b)}{\Gamma(d)}\right)^s}\enspace$ .

$(B)\hspace{1cm}$ Series expansion.

It’s $\enspace\displaystyle\frac{\Gamma(d,x)}{\Gamma(d)}=\frac{1}{e^x}\sum\limits_{k=0}^{d-1}\frac{x^k}{k!}\enspace$ .

If we define $\enspace\displaystyle \sum\limits_{k=0}^{mn} a_{m,n,k}x^k := \left(\sum\limits_{k=0}^m \frac{x^k}{k!}\right)^n \enspace$ then we can calculate explicit

$\displaystyle \int\limits_0^\infty x^d (b+cx)^{ds} \left(1-\frac{\Gamma(d,b+cx)}{\Gamma(d)}\right)^{-s} e^{-asx}dx = $

$\displaystyle =\sum\limits_{n=0}^\infty {\binom {s+n-1}{n}} \frac{1}{e^{bn}} \sum\limits_{k=0}^{(d-1)n} a_{d-1,n,k} \sum\limits_{j=0}^\infty {\binom {ds+k}{j}} \frac{b^{ds+k-j} c^j (d+j)!}{(as+cn)^{d+j+1}}\enspace$ .

It makes sense to choose natural numbers for $\,s\,$ .

  • $\begingroup$ I don't really follow how you go from the integral to the series representation, is it simply an application of the fractional binomial theorem? Also, where do you use the assumption of cd<ab in the series expansion? $\endgroup$
    – Johan
    Sep 19, 2017 at 15:28
  • $\begingroup$ You are right, $cd<ab$ belongs to $(A)$, I will correct it . And Yes: Simply the binomial theorem and the integral for the Gamma function, nothing special. $\endgroup$
    – user90369
    Sep 19, 2017 at 15:37
  • $\begingroup$ Alright, I guess this series representation is as good as it gets. Thanks! $\endgroup$
    – Johan
    Sep 19, 2017 at 15:38
  • $\begingroup$ You are welcome. :-) $\endgroup$
    – user90369
    Sep 19, 2017 at 15:41

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