So here is my question,
I would like to compute the following limit, $$\lim_{n\rightarrow\infty}\int_{\mathbb R_+}\frac{sin(x)}{x}(e^{-x/n}-1)dx$$ To interchange the integral an the limit I want to apply Dominated Convergence i.e i have to bound $|\frac{sin(x)}{x}(e^{-x/n}-1)|$ with some integrable function. I know I can obtain such a bound by integrating $\frac{sin(x)}{x}(e^{-x/n}-1)$ by parts. So I wanted to ask if someone knows a direct way to bound without using integration by parts? Thanks.
Edit
This is a part of an exercise where the goal is to prove that $\int_{\mathbb R_+}\frac{sin(x)}{x}=\pi/2$. So i am not allowed to use that $\int_{\mathbb R_+}\frac{sin(x)}{x}$ is finite.
Moreover in the solution of the exercise, where the one goal is to show that, $$\lim_{n\rightarrow\infty}|\int_{\mathbb R_+}\frac{sin(x)}{x}(e^{-x/n}-1)dx|=0$$ My professor wrote this to prove the above statement,
An integration by parts shows that, $$|\int_{\mathbb R_+}\frac{sin(x)}{x}(e^{-x/n}-1)dx|=|[\frac{1-cos(x)}{x}(e^{-x/n}-1)]_0^{\infty}-\int_{\mathbb R_+}\frac{1-cos(x)}{x^2}[1-e^{-x/n}(1+x/n)]dx|\leq|\int_{\mathbb R_+}\frac{1-cos(x)}{x^2}[1-e^{-x/n}(1+x/n)]dx|$$ Then he defines $g(x)_n:=\frac{1-cos(x)}{x^2}|1-e^{-x/n}(1+x/n)|$. Clearly $\lim_{n\rightarrow\infty}g(x)_n=0$. Furthermore since $g(x)_n\leq\frac{1-cos(x)}{x^2}$ and $\frac{1-cos(x)}{x^2}$ is integrable we can finally conclude, $$|\frac{sin(x)}{x}(e^{-x/n}-1)|\leq g(x)_n\rightarrow0$$ what proves the upper claim.
So i wanted to know if there is an easier way to obtain this?
You cannot bound the non-negative functions $$ f(x)=\frac{1-e^{-x/n}}{x}\cdot\left|\sin x\right| $$ or (it is clearly the same) $$ g_n(x) = \frac{1-e^{-x}}{x}\cdot\left|\sin(nx)\right| $$ with an integrable function over $\mathbb{R}^+$, since there exists a translation-invariant subset $H$ of $\mathbb{R}^+$ with infinite measure where $|\sin(n x)|\geq\frac{1}{2}$. Over $K=H\cap\{x\in\mathbb{R}:x>\log 2\}$ $$ g_{n}(x) > \frac{1}{4x} $$ holds, while $\frac{1}{4x}$ is not an integrable function over $K$.