Integral of incomplete gamma function and limit of hypergeometric function 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?
 A: $$\begin{eqnarray*} \int_{x}^{+\infty}\frac{\Gamma(a,t)}{t}\,dt = \int_{1}^{+\infty}\frac{\Gamma(a,sx)}{s}\,ds&=&\int_{1}^{+\infty}\int_{sx}^{+\infty}\frac{1}{s}u^{a-1}e^{-u}\,du\,ds\\&=&x^a\iint_{(1,+\infty)^2}(sv)^{a-1}e^{-svx}\,dv\,ds\\&=&x^a\int_{1}^{+\infty}\int_{1}^{p}\frac{1}{s} p^{a-1}e^{-px}\,ds\,dp\\&=&x^a\int_{1}^{+\infty}\log(p) p^{a-1} e^{-px}\,dp\\&=&x^a\cdot\frac{d}{da}\int_{1}^{+\infty}p^{a-1}e^{-px}\,dx\\&=&x^a\cdot\color{purple}{\frac{d}{da}}\left(\color{red}{x^{-a}\Gamma(a)}-\color{blue}{\int_{0}^{1}p^{a-1}e^{-px}\,dp}\right)\tag{1}\end{eqnarray*} $$
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:

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

