Prove the limit equation: $\lim_{x\rightarrow\infty}e^{-x}\sum_{n=0}^{\infty}f_{n}\left(x\right)=-\gamma.$ FYI$,\quad\gamma$ is Euler's constant Define a function sequence $\left \{ f_{n} \right \}_{n= 0}^{\infty}$ on $\left ( 0, \infty \right )$ as
$$f_{0}\left ( x \right )= \ln x,\quad f_{n+ 1}\left ( x \right )= \int_{0}^{x}f_{n}\left ( t \right ){\rm d}t\quad\left ( n= 0, 1, 2\cdots \right )$$
Prove the limiting equation:
$$\lim_{x\rightarrow\infty}e^{-x}\sum_{n= 0}^{\infty}f_{n}\left ( x \right )= -\gamma$$
For your information$,\quad\gamma$ is Euler's constant defined
$$\gamma:=\lim\left ( \sum_{k= 1}^{n}\frac{1}{k}- \ln n \right )$$
This problem belongs to my friend@twelve_sakuya, I haven't touched Euler's constant of course we have no approaches from me. But I had done a similar work. Maybe, there is a relationship between both problems, I need to the help. Here we go.

Given a function $f\left ( x \right )$ continuous on the real interval. Prove that
$$\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$
Source: Hungarian Math Contest Revision

Integral's limit, which I think that required us to divide and evaluate the integral function. By using the integral form of Cauchy-Schwarz inequality, we have
$$\left | e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq e^{-x}\left | \int_{-x}^{-\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{\frac{x}{2}}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq$$
$$\leq e^{-x}\left ( \int_{-x}^{-\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-x}^{-\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+ e^{-x}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+$$
$$+ e^{-x}\left ( \int_{\frac{x}{2}}^{x}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{\frac{x}{2}}^{x}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}\leq e^{-\frac{3}{2}x}\left ( \int_{-\infty}^{-\frac{x}{2}}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+ e^{-\frac{x}{2}}\left ( \int_{-\infty}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+$$
$$+ \left ( \int_{\frac{x}{2}}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}\underset{x\rightarrow\infty}{\rightarrow}0\therefore\left | \int_{-\infty}^{\infty}f^{2}\left ( x \right ){\rm d}x \right |< \infty\Rightarrow\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$
 A: This is not an answer, it's more like a hint, but it's too large for a comment. The $n$-th term of the sequence $\{f_n\}$ can be written as iterated integrals. For example
$$f_2 = \int_{0}^{x}\int_{0}^{t}\log u\,du \,dt.$$
The Cauchy formula for repeated integration states that if
$$I^n(x) = \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1,$$
this expression is equal to
$$I^n(x) = \frac{1}{(n-1)!} \int_a^x \left(x-t\right)^{n-1} f(t)\,\mathrm{d}t.$$
Edit: I saw Yuris comment just now. Not a copy.
A: $$f_n(x)=\frac{1}{(n-1)!}\int_0^x (x-t)^{n-1} \ln t ~dt$$

$$\sum_{n= 1}^{\infty}f_{n}\left ( x \right )=\int_0^x \ln t ~dt~\sum_{n= 1}^{\infty} \frac{(x-t)^{n-1}}{(n-1)!}=\int_0^x \ln t ~dt~e^{x-t}$$

$$\lim_{x \to \infty} e^{-x}\sum_{n= 1}^{\infty}f_{n}=\int_0^\infty e^{-t} \ln t ~dt=-\gamma$$
The summation could be started with $1$ and not with $0$, as $e^{-x} f_0(x)$ simply gives zero in the limit.
A: Induction shows that
$$
f_n(x)=\frac{x^n}{n!}(\log(x)-H_n)\tag1
$$
where $H_n$ are the Harmonic Numbers.
Define
$$
s(x)=\sum_{n=0}^\infty f_n(x)\tag2
$$
$(1)$ shows that the sum in $(2)$ converges for all $x\ne0$.
Compute the derivative:
$$
\begin{align}
s'(x)
&=\sum_{n=0}^\infty f_n'(x)\tag{3a}\\
&=\frac1x+\sum_{n=1}^\infty f_n'(x)\tag{3b}\\
&=\frac1x+\sum_{n=0}^\infty f_n(x)\tag{3c}\\
&=\frac1x+s(x)\tag{3d}
\end{align}
$$
Rearranging $(3)$, multiplying by the integrating factor $e^{-x}$, and subtracting $\frac1x$ yields
$$
\left(e^{-x}s(x)-\log(x)\right)'=\frac{e^{-x}-1}x\tag4
$$
Integration by parts gives
$$
\begin{align}
e^{-x}s(x)
&=\log(x)+\int_0^x\frac{e^{-t}-1}t\mathrm{d}t\tag{5a}\\
&=\log(x)\,e^{-x}+\int_0^x\log(t)\,e^{-t}\,\mathrm{d}t\tag{5b}
\end{align}
$$
As shown in this answer,
$$
\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=-\gamma\tag6
$$
Applying $(6)$ to the limit of $(5)$ gives
$$
\lim_{x\to\infty}e^{-x}s(x)=-\gamma\tag7
$$
A: First of all, the series $\sum_{n\geqslant 0}f_n(x)$ converges for all $x>0$ because one can show by induction that $|f_n(x)|\leqslant\frac{(x+1)^{n+1}}{(n+1)!}$. The case $n=0$ being straightforward, let $n$ such that $|f_n(x)|\leqslant\frac{(x+1)^{n+1}}{(n+1)!}$ for all $x>0$, then
$$ |f_{n+1}(x)|\leqslant\frac{1}{(n+1)!}\int_0^x (t+1)^{n+1}dt=\frac{(x+1)^{n+2}}{(n+2)!} $$
Moreover, it follows that $\sum_{n\geqslant 1} f_n'=\sum_{n\geqslant 0}f_n$ is normally convergent on all compact of $(0,+\infty)$, therefore, if $\varphi:=\sum_{n\geqslant}f_n$, then $\varphi$ is differentiable and
$$ \varphi'(x)=\sum_{n\geqslant 0}f_n'(x)=\sum_{n\geqslant 0}f_n(x)+f_0'(x)=\varphi(x)+\frac{1}{x} $$
Thus there exists $\lambda\in\mathbb{R}$ such that $\displaystyle\varphi(x)=\lambda e^x+\int_1^x\frac{e^{x-t}}{t}dt$ so that
$$ \lim\limits_{x\rightarrow +\infty}\varphi(x)e^{-x}=\lambda+\int_1^{+\infty}\frac{e^{-t}}{t}dt $$
But since $f_n(0)=0$ for all $n\geqslant 1$, then $\lim\limits_{x\rightarrow 0}(\varphi(x)-\log x)=0$. Using an integration by parts, we have
$$ \int_1^x\frac{e^{x-t}}{t}dt=\log x+\int_1^x e^{x-t}\log tdt $$
Therefore $\lambda=\int_0^1e^{-t}\log tdt$. Finally,
$$ \lim\limits_{x\rightarrow +\infty}\varphi(x)e^{-x}=\int_0^{+\infty}e^{-t}\log tdt=\Gamma'(1)=-\gamma $$
