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Let $a > 0$ and consider the integral $$\int_x^\infty \frac{\Gamma(a,t)}{t}\,dt$$ where $\Gamma(a,t)$ is the upper incomplete gamma function $$\Gamma(a,t) = \int_t^\infty x^{a-1} e^{-x} \, dx.$$ Wolfram Alpha says that this definite integral equals $$\Gamma'(a) - \Gamma(a) \log(x) + \frac{x^a}{a^2} {}_2 F_2(a,a;a+1,a+1;-x)$$ where ${}_2 F_2$ is the hypergeometric function which has series representation $${}_2 F_2(a,a;a+1,a+1;-x) = \sum_{k=0}^\infty \frac{a^2}{k! (a+k)^2} (-x)^k.$$ The indefinite integral is clear enough by doing term-by-term integration. But for this definite integral to be true, we must also prove the limit $$\lim_{x \rightarrow \infty} \Gamma(a) \log(x) - \frac{x^a}{a^2} {}_2 F_2(a,a;a+1,a+1;-x) = \Gamma'(a).$$

How to prove this limit?

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  • $\begingroup$ my bad, it is always $\Gamma(a,t) = \int_t^\infty x^{a-1} e^{-x} dx$ en.wikipedia.org/wiki/Incomplete_gamma_function $\endgroup$ Commented Jun 3, 2016 at 21:47
  • $\begingroup$ then it is obvious that $\int_x^\infty \Gamma(a,t)t^k dt$ converges for any $k$, since for $t$ large enough : $\Gamma(a,t) < \int_t^\infty e^{-x/2} dx = \mathcal{O}(e^{-t/2})$, this is why your $ _2F_2$ series representation is entire in $x$. and it shows that $\int_x^\infty \frac{\Gamma(a,t)}{t} dt \to 0$ as $x \to \infty$ $\endgroup$ Commented Jun 3, 2016 at 21:54
  • $\begingroup$ Yes that is clear. But that does not explain the particular value of the limit, does it? $\endgroup$ Commented Jun 3, 2016 at 22:07

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$$\begin{eqnarray*} \int_{x}^{+\infty}\frac{\Gamma(a,t)}{t}\,dt = \int_{1}^{+\infty}\frac{\Gamma(a,sx)}{s}\,ds&=&\int_{1}^{+\infty}\int_{sx}^{+\infty}\frac{1}{s}u^{a-1}e^{-u}\,du\,ds\\&=&x^a\iint_{(1,+\infty)^2}(sv)^{a-1}e^{-svx}\,dv\,ds\\&=&x^a\int_{1}^{+\infty}\int_{1}^{p}\frac{1}{s} p^{a-1}e^{-px}\,ds\,dp\\&=&x^a\int_{1}^{+\infty}\log(p) p^{a-1} e^{-px}\,dp\\&=&x^a\cdot\frac{d}{da}\int_{1}^{+\infty}p^{a-1}e^{-px}\,dx\\&=&x^a\cdot\color{purple}{\frac{d}{da}}\left(\color{red}{x^{-a}\Gamma(a)}-\color{blue}{\int_{0}^{1}p^{a-1}e^{-px}\,dp}\right)\tag{1}\end{eqnarray*} $$ but by expanding $e^{-px}$ as a Taylor series: $$ \color{blue}{\int_{0}^{1}p^{a-1}e^{-px}\,dp}=\sum_{n\geq 0}\frac{(-1)^n\,x^n}{n!}\int_{0}^{1}p^{n+a-1}\,dp =\sum_{n\geq 0}\frac{(-1)^n x^n}{n!(n+a)}\tag{2}$$ hence: $$\color{purple}{\frac{d}{da}}\color{blue}{\int_{0}^{1}p^{a-1}e^{-px}\,dp}=-\sum_{n\geq 0}\frac{(-1)^n x^n}{n!(n+a)^2}\tag{3}$$ while: $$\color{purple}{\frac{d}{da}}\color{red}{x^{-a}\Gamma(a)} = x^{-a}\,\Gamma'(a)-\log(x)\, x^{-a}\,\Gamma(a)\tag{4}$$ and the proof is complete:

$$ \int_{x}^{+\infty}\frac{\Gamma(a,t)}{t}\,dt = \color{purple}{\Gamma'(a)-\Gamma(a)\log(x)+\sum_{n\geq 0}\frac{(-1)^n x^{n+a}}{n!(n+a)^2}}.\tag{5}$$

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  • $\begingroup$ What a beautiful proof! Please tell me: How did you find it? $\endgroup$ Commented Jun 5, 2016 at 3:38
  • $\begingroup$ @JohnM: I recognized a hypergeometric-like structure in the original double integral, then tried to isolate the singular part from the regular part through some substitutions, and it worked quite nice in few tries. $\endgroup$ Commented Jun 5, 2016 at 13:56

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