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

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

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

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

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

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

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