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$\newcommand{\Ga}{\Gamma} $Let us show that
\begin{equation*} T(4,1,x)\sim t(x):=e^{-x}/x^3 \tag{1}\label{1} \end{equation*} as $x\to\infty$.

Indeed, here are formulas (29) and (30) of the linked paper: \begin{equation*} \frac{d\Ga(a,x)}{da}=\ln x\,\Ga(a,x)+x\,T(3,a,x), \tag{29}\label{29} \end{equation*} \begin{equation*} \frac{d^2\Ga(a,x)}{da^2}=\ln^2 x\,\Ga(a,x)+2x(\ln x\,T(3,a,x)+T(4,a,x)). \tag{30}\label{30} \end{equation*}

Eliminating $T(3,a,x)$ from these equations and noting that \begin{equation*} \frac{d^k\Ga(a,x)}{da^k}=\int_x^\infty u^{a-1}\ln^k u\,e^{-u}\,du \end{equation*} for natural $k$, we get $$T(4,1,x)=\frac{g_2(x)-2 g_1(x) \ln x+e^{-x} \ln^2 x}{2x},$$ where $$g_k(x):=\int_x^\infty\ln^k u\,e^{-u}\,du.$$ Using now repeatedly integration by parts to obtain enough terms of the asymptotics of $g_1(x)$ and $g_2(x)$, we get \eqref{1} -- see details on this at the end of this answer.

For an illustration, here is the graph $\{(x,\frac{T(4,1,x)}{t(x)})\colon20\le x\le200\}$:

Details: Let $l$ denote a function of the form $\ln^p$ for some real $p$ or any of its derivatives, of any order. Integrating by parts, for all real $x>1$ we have \begin{equation*} g(x):=\int_x^\infty l(u) e^{-u}\,du=l(x)e^{-x}+\int_x^\infty l'(u) e^{-u}\,du \end{equation*} and hence, by induction, for any natural $k$, \begin{equation*} \begin{aligned} g(x)&=(l(x)+l'(x)+\cdots+l^{(k-1)}(x))e^{-x}+\int_x^\infty l^{(k)}(u) e^{-u}\,du \\ &=(l(x)+l'(x)+\cdots+l^{(k-1)}(x)+l^{(k)}(x)(1+o(1)))e^{-x} \end{aligned} \tag{2}\label{2} \end{equation*} (as $x\to\infty$); the latter equality in \eqref{2} follows by the l'Hospital rule, since $(\ln^p)^{(k)}(x)\asymp x^{-k}\ln^{p-1}x$.

So, by \eqref{2} with $k=3$ and $g\in\{g_1,g_2\}$, \begin{equation*} e^x g_1(x)=\ln x+\frac{1}{x}-\frac{1}{x^2}+\frac{2+o(1)}{x^3}, \end{equation*} \begin{equation*} e^x g_2(x)=\ln^2 x+\frac{2 \ln x}{x}-\frac{2 \ln x}{x^2}+\frac{2}{x^2} +\frac{(4+o(1)) \ln x}{x^3}, \end{equation*} \begin{equation*} e^x T(4,1,x)=\frac{x+o(\ln x)}{x^4}\sim 1/x^3. \quad\Box \end{equation*}

$\newcommand{\Ga}{\Gamma} $We have \begin{equation*} T(4,1,x)\sim t(x):=e^{-x}/x^3 \tag{1}\label{1} \end{equation*} as $x\to\infty$.

Indeed, from formulas (29) and (30) of the linked paper we get $$T(4,1,x)=\frac{g_2(x)-2 g_1(x) \ln x+e^{-x} \ln^2 x}{2 x},$$ where $$g_k(x):=\int_x^\infty\ln^k u\,e^{-u}\,du.$$ Using now repeatedly integration by parts to obtain enough terms of the asymptotics of $g_1(x)$ and $g_2(x)$, we get \eqref{1} -- see details on this at the end of this answer.

Here is the graph $\{(x,\frac{T(4,1,x)}{t(x)})\colon20\le x\le200\}$:

Details: Let $l$ denote a function of the form $\ln^p$ for some real $p$ or any of its derivatives, of any order. Integrating by parts, for all real $x>1$ we have \begin{equation*} g(x):=\int_x^\infty l(u) e^{-u}\,du=l(x)e^{-x}+\int_x^\infty l'(u) e^{-u}\,du \end{equation*} and hence, by induction, for any natural $k$, \begin{equation*} \begin{aligned} g(x)&=(l(x)+l'(x)+\cdots+l^{(k-1)}(x))e^{-x}+\int_x^\infty l^{(k)}(u) e^{-u}\,du \\ &=(l(x)+l'(x)+\cdots+l^{(k-1)}(x)+l^{(k)}(x)(1+o(1)))e^{-x} \end{aligned} \tag{2}\label{2} \end{equation*} (as $x\to\infty$). The latter equality in \eqref{2} follows by the l'Hospital rule, because $(\ln^p)^{(k)}(x)\asymp x^{-k}\ln^{p-1}x$ and hence $r(x):=l^{(k+1)}(x)/l^{(k)}(x)\to0$. $\Big($Indeed, \begin{equation*} \frac{\Big(\int_x^\infty l^{(k)}(u) e^{-u}\,du\Big)'_x} {\big(l^{(k)}(x) e^{-x}\big)'_x}\ =\frac1{1-r(x)}\to1, \end{equation*} where ${}'_x$ denotes the differentiation in $x$.$\Big)$

So, by \eqref{2} with $k=3$ and $g\in\{g_1,g_2\}$, \begin{equation*} e^x g_1(x)=\ln x+\frac{1}{x}-\frac{1}{x^2}+\frac{2+o(1)}{x^3}, \end{equation*} \begin{equation*} e^x g_2(x)=\ln^2 x+\frac{2 \ln x}{x}-\frac{2 \ln x}{x^2}+\frac{2}{x^2} +\frac{(4+o(1)) \ln x}{x^3}, \end{equation*} \begin{equation*} e^x T(4,1,x)=\frac{x+o(\ln x)}{x^4}\sim 1/x^3. \quad\Box \end{equation*}

We have \begin{equation*} T(4,1,x)\sim t(x):=e^{-x}/x^3 \tag{1}\label{1} \end{equation*} as $x\to\infty$.

Indeed, from formulas (29) and (30) of the linked paper we get $$T(4,1,x)=\frac{g_2(x)-2 g_1(x) \ln x+e^{-x} \ln^2 x}{2 x},$$ where $$g_k(x):=\int_x^\infty\ln^k u\,e^{-u}\,du.$$ Using now repeatedly integration by parts to obtain enough terms of the asymptotics of $g_1(x)$ and $g_2(x)$, we get \eqref{1} -- see details on this at the end of this answer.

Here is the graph $\{(x,\frac{T(4,1,x)}{t(x)})\colon20\le x\le200\}$:

Details: Let $l$ denote a function of the form $\ln^p$ for some real $p$ or any of its derivatives, of any order. Integrating by parts, for all real $x>1$ we have \begin{equation*} g(x):=\int_x^\infty l(u) e^{-u}\,du=l(x)e^{-x}+\int_x^\infty l'(u) e^{-u}\,du \end{equation*} and hence, by induction, for any natural $k$, \begin{equation*} \begin{aligned} g(x)&=(l(x)+l'(x)+\cdots+l^{(k-1)}(x))e^{-x}+\int_x^\infty l^{(k)}(u) e^{-u}\,du \\ &=(l(x)+l'(x)+\cdots+l^{(k-1)}(x)+l^{(k)}(x)(1+o(1)))e^{-x} \end{aligned} \tag{2}\label{2} \end{equation*} (as $x\to\infty$); the latter equality in \eqref{2} follows by the l'Hospital rule, since $(\ln^p)^{(k)}(x)\asymp x^{-k}\ln^{p-1}x$.

So, by \eqref{2} with $k=3$ and $g\in\{g_1,g_2\}$, \begin{equation*} e^x g_1(x)=\ln x+\frac{1}{x}-\frac{1}{x^2}+\frac{2+o(1)}{x^3}, \end{equation*} \begin{equation*} e^x g_2(x)=\ln^2 x+\frac{2 \ln x}{x}-\frac{2 \ln x}{x^2}+\frac{2}{x^2} +\frac{(4+o(1)) \ln x}{x^3}, \end{equation*} \begin{equation*} e^x T(4,1,x)=\frac{x+o(\ln x)}{x^4}\sim 1/x^3. \quad\Box \end{equation*}

$\newcommand{\Ga}{\Gamma} $Let us show that
\begin{equation*} T(4,1,x)\sim t(x):=e^{-x}/x^3 \tag{1}\label{1} \end{equation*} as $x\to\infty$.

Indeed, here are formulas (29) and (30) of the linked paper: \begin{equation*} \frac{d\Ga(a,x)}{da}=\ln x\,\Ga(a,x)+x\,T(3,a,x), \tag{29}\label{29} \end{equation*} \begin{equation*} \frac{d^2\Ga(a,x)}{da^2}=\ln^2 x\,\Ga(a,x)+2x(\ln x\,T(3,a,x)+T(4,a,x)). \tag{30}\label{30} \end{equation*}

Eliminating $T(3,a,x)$ from these equations and noting that \begin{equation*} \frac{d^k\Ga(a,x)}{da^k}=\int_x^\infty u^{a-1}\ln^k u\,e^{-u}\,du \end{equation*} for natural $k$, we get $$T(4,1,x)=\frac{g_2(x)-2 g_1(x) \ln x+e^{-x} \ln^2 x}{2x},$$ where $$g_k(x):=\int_x^\infty\ln^k u\,e^{-u}\,du.$$ Using now repeatedly integration by parts to obtain enough terms of the asymptotics of $g_1(x)$ and $g_2(x)$, we get \eqref{1} -- see details on this at the end of this answer.

For an illustration, here is the graph $\{(x,\frac{T(4,1,x)}{t(x)})\colon20\le x\le200\}$:

Details: Let $l$ denote a function of the form $\ln^p$ for some real $p$ or any of its derivatives, of any order. Integrating by parts, for all real $x>1$ we have \begin{equation*} g(x):=\int_x^\infty l(u) e^{-u}\,du=l(x)e^{-x}+\int_x^\infty l'(u) e^{-u}\,du \end{equation*} and hence, by induction, for any natural $k$, \begin{equation*} \begin{aligned} g(x)&=(l(x)+l'(x)+\cdots+l^{(k-1)}(x))e^{-x}+\int_x^\infty l^{(k)}(u) e^{-u}\,du \\ &=(l(x)+l'(x)+\cdots+l^{(k-1)}(x)+l^{(k)}(x)(1+o(1)))e^{-x} \end{aligned} \tag{2}\label{2} \end{equation*} (as $x\to\infty$); the latter equality in \eqref{2} follows by the l'Hospital rule, since $(\ln^p)^{(k)}(x)\asymp x^{-k}\ln^{p-1}x$.

So, by \eqref{2} with $k=3$ and $g\in\{g_1,g_2\}$, \begin{equation*} e^x g_1(x)=\ln x+\frac{1}{x}-\frac{1}{x^2}+\frac{2+o(1)}{x^3}, \end{equation*} \begin{equation*} e^x g_2(x)=\ln^2 x+\frac{2 \ln x}{x}-\frac{2 \ln x}{x^2}+\frac{2}{x^2} +\frac{(4+o(1)) \ln x}{x^3}, \end{equation*} \begin{equation*} e^x T(4,1,x)=\frac{x+o(\ln x)}{x^4}\sim 1/x^3. \quad\Box \end{equation*}

We have \begin{equation*} T(4,1,x)\sim t(x):=e^{-x}/x^3 \tag{1}\label{1} \end{equation*} as $x\to\infty$.

Indeed, from formulas (29) and (30) of the linked paper we get $$T(4,1,x)=\frac{g_2(x)-2 g_1(x) \ln x+e^{-x} \ln^2 x}{2 x},$$ where $$g_k(x):=\int_x^\infty\ln^k u\,e^{-u}\,du.$$ Using now repeatedly integration by parts to obtain enough terms of the asymptotics of $g_1(x)$ and $g_2(x)$, we get \eqref{1} -- see details on this at the end of this answer.

Here is the graph $\{(x,\frac{T(4,1,x)}{t(x)})\colon20\le x\le200\}$:

Details: Let $l$ denote a function of the form $\ln^p$ for some real $p$ or any of its derivatives, of any order. Integrating by parts, for all real $x>0$ we have \begin{equation*} g(x):=\int_x^\infty l(u) e^{-u}\,du=l(x)e^{-x}+\int_x^\infty l'(u) e^{-u}\,du \end{equation*} and hence, by induction, for any natural $k$, \begin{equation*} \begin{aligned} g(x)&=(l(x)+l'(x)+\cdots+l^{(k-1)}(x))e^{-x}+\int_x^\infty l^{(k)}(u) e^{-u}\,du \\ &=(l(x)+l'(x)+\cdots+l^{(k-1)}(x)+l^{(k)}(x)(1+o(1)))e^{-x} \end{aligned} \tag{2}\label{2} \end{equation*} (as $x\to\infty$); the latter equality in \eqref{2} follows by the l'Hospital rule, since $(\ln^p)^{(k)}(x)\asymp x^{-k}\ln^{p-1}x$.

So, by \eqref{2} with $k=3$ and $g\in\{g_1,g_2\}$, \begin{equation*} e^x g_1(x)=\ln x+\frac{1}{x}-\frac{1}{x^2}+\frac{2+o(1)}{x^3}, \end{equation*} \begin{equation*} e^x g_2(x)=\ln^2 x+\frac{2 \ln x}{x}-\frac{2 \ln x}{x^2}+\frac{2}{x^2} +\frac{(4+o(1)) \ln x}{x^3}, \end{equation*} \begin{equation*} e^x T(4,1,x)=\frac{x+o(\ln x)}{x^4}\sim 1/x^3. \quad\Box \end{equation*}

We have \begin{equation*} T(4,1,x)\sim t(x):=e^{-x}/x^3 \tag{1}\label{1} \end{equation*} as $x\to\infty$.

Indeed, from formulas (29) and (30) of the linked paper we get $$T(4,1,x)=\frac{g_2(x)-2 g_1(x) \ln x+e^{-x} \ln^2 x}{2 x},$$ where $$g_k(x):=\int_x^\infty\ln^k u\,e^{-u}\,du.$$ Using now repeatedly integration by parts to obtain enough terms of the asymptotics of $g_1(x)$ and $g_2(x)$, we get \eqref{1} -- see details on this at the end of this answer.

Here is the graph $\{(x,\frac{T(4,1,x)}{t(x)})\colon20\le x\le200\}$:

Details: Let $l$ denote a function of the form $\ln^p$ for some real $p$ or any of its derivatives, of any order. Integrating by parts, for all real $x>1$ we have \begin{equation*} g(x):=\int_x^\infty l(u) e^{-u}\,du=l(x)e^{-x}+\int_x^\infty l'(u) e^{-u}\,du \end{equation*} and hence, by induction, for any natural $k$, \begin{equation*} \begin{aligned} g(x)&=(l(x)+l'(x)+\cdots+l^{(k-1)}(x))e^{-x}+\int_x^\infty l^{(k)}(u) e^{-u}\,du \\ &=(l(x)+l'(x)+\cdots+l^{(k-1)}(x)+l^{(k)}(x)(1+o(1)))e^{-x} \end{aligned} \tag{2}\label{2} \end{equation*} (as $x\to\infty$); the latter equality in \eqref{2} follows by the l'Hospital rule, since $(\ln^p)^{(k)}(x)\asymp x^{-k}\ln^{p-1}x$.

So, by \eqref{2} with $k=3$ and $g\in\{g_1,g_2\}$, \begin{equation*} e^x g_1(x)=\ln x+\frac{1}{x}-\frac{1}{x^2}+\frac{2+o(1)}{x^3}, \end{equation*} \begin{equation*} e^x g_2(x)=\ln^2 x+\frac{2 \ln x}{x}-\frac{2 \ln x}{x^2}+\frac{2}{x^2} +\frac{(4+o(1)) \ln x}{x^3}, \end{equation*} \begin{equation*} e^x T(4,1,x)=\frac{x+o(\ln x)}{x^4}\sim 1/x^3. \quad\Box \end{equation*}