For $n\in \mathbb N$, consider the real functions
$$f_n(x):=\int_2^x\dfrac{dt}{t^ne^t\log t}.$$
(1) Find the domain of the functions
(2) Show that $f(x)=\sum_{n=2}^{+\infty}f_n$ is $\mathcal C^1$ on $(1,+\infty).$
(3) Is $\underset{n\to+\infty}\lim f(x)$ finite?
Let $g_n(x)=\dfrac{1}{x^ne^x\log x}$ be the integrand function $\implies g_n(x)=\dfrac{1}{e^x\log x}\Big(\dfrac{1}{x}\Big)^n$ is defined in $(0,+\infty)-\{1\}$. If I now fix a real number $x_0\in \mathbb R$ with $n\to +\infty$, the functions $g_n(x)\not\to\infty\iff -1<\dfrac{1}{x}< 1\implies x>1$.
So the interval in which we have punctual and also uniform (I'd like to prove this more precisely) convergence is $(1,+\infty)$ .
If I now want to study the differentiability of $f(x)$ I considered the derivative of the general term $\int_2^x\frac{dt}{t^ne^t\log t}$ in order to find a limit function $h(x)$ such that $f'_n(x)\to h(x)$ uniformly.
$$\dfrac{d}{dx}\int_2^x\dfrac{dt}{t^ne^t\log t}=\dfrac{1}{x^ne^x\log x}\underset{\text{uniformly}}\longrightarrow h(x)\equiv 0\text{ in } (1,+\infty)\implies$$
the series $\sum f_n$ is differentiable in $(1,+\infty)$.
For the third point I'm not sure about how to proceed...
Let $1<x<\infty$. We can write $f(x)= \sum\limits_{n=2}^{\infty}\int_2^{x} \frac 1 {t^{n}e^{t} \ln t}dt= \int_2^{x} \sum\limits_{n=2}^{\infty}\frac 1 {t^{n}e^{t} \ln t}dt$ since the interchange of sum and intetgral is justified by the non-negativity of the integrand. Using the formula for a geometric sum we get $f(x)=\int_2^{x} \frac 1 {t(t-1) e^{t}\ln t}dt$. By FTC this is differentiable on $(1,\infty)$ and $f'(x)=\frac 1 {x(x-1) e^{x}\ln x}$ and this derivative is continuous on that interval. Finally, the fact that $0< \frac 1 {t(t-1) e^{t}\ln t}<Ce^{-t}$ for some constant $C$ for lal $t>2$ shows that $\lim_{x \to \infty} f(x)$ exists and equals $\int_2^{\infty} \frac 1 {t(t-1) e^{t}\ln t}dt$.