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. 

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

[![enter image description here][1]][1]

---

**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*}

  [1]: https://i.sstatic.net/cQDHA.png