I want to prove that: $$ f_n: (0,\infty) \to \mathbb{R}, \quad f_n(x) =\frac{\log(e^{nx} + \sin{x})}{nx(1 + x)^2}$$
is in $L^1([0,\infty))$ for all $n \ge 1$
My source suggests to split the reasoning in 2 steps:
- $f_n \in L^1([0,1])$
- $f_n \in L^1([1,\infty))$
For the first step I'd like to use $\log(1 + x) \le x$ and $\sin x \le x$ but I have no idea how to do it.
I'd like to find a dominating function in order to, later, use the dominated convergence theorem.
Any suggestions on how to carry out those inequalities?
Write $\log\left(e^{nx}+\sin x\right)=nx+\log\left(1+e^{-nx} \sin x \right)$ in order to get $$f_n(x)=\frac 1{(1+x)^2} +g_n(x),\mbox{ with }g_n(x):=\frac{\log\left(1+e^{-nx} \sin x \right)}{nx \left(1+x\right)^2}.$$ It suffices to establish integrability of $g_n$.
When $x$ is large enough (say than $x_0$, we have $1/2 \lt 1+e^{-nx} \sin x\lt 3/2$ and the function $x\mapsto 1/(x(1+x)^2)$ is integrable on $[x_0, +\infty)$. There is no problem on $[1,x_0]$, so we are reduced to prove that $g_n\in\mathbb L^1\left([0,1]\right)$. The only problem is at $0$, but $x\mapsto \log\left( 1+e^{-nx} \sin x \right) /x$ has a limit as $x$ goes to $0$ (the derivative of the function $x\mapsto \log\left( 1+e^{-nx} \sin x \right)$ at $0$.