$f \in L^1(- \infty , + \infty)$ , Find the limit $ \lim_{n \to \infty} \int^{{+ \infty}}_{- \infty} \frac{f(x)e^{nx}}{1+e^{nx}} dx $
My Attempt
$$ \lim_{n \to \infty} \int^{{+ \infty}}_{- \infty} \frac{f(x)e^{nx}}{1+e^{nx}} dx = \lim_{n \to \infty} \int^{{+ \infty}}_{- \infty} \frac{f(x)}{1+e^{-nx}} dx $$
Define $g_n(x)= \frac{f(x)}{1+e^{-nx}}$, The integrable function $f(x)$ dominates all $g_n(x)$ for all $x \in (-\infty, + \infty)$ due to the simple fact that $1+e^{-nx} > 1$ for all $n$ and all $x \in (-\infty, + \infty)$. Now using the Lebesgue's Dominated Convergence theorem we have $$ \begin{align} \lim_{n \to \infty} \int^{{+ \infty}}_{- \infty} \frac{f(x)}{1+e^{-nx}} dx &= \int^{{+ \infty}}_{- \infty} \lim_{n \to \infty} \frac{f(x)}{1+e^{-nx}} dx \\ &= \int^{0}_{- \infty} \lim_{n \to \infty} \frac{f(x)}{1+e^{-nx}} dx + \int^{{+ \infty}}_{0} \lim_{n \to \infty} \frac{f(x)}{1+e^{-nx}} dx \end{align} $$ Since $f \in L^1(- \infty , + \infty)$ we conclude that $f$ is finite a.e. on $(- \infty , + \infty)$ and hence $$\lim_{n \to \infty} \frac{f(x)}{1+e^{-nx}} = 0 \ \ \ \ if \ \ x\in (-\infty,0)$$ and $$ \lim_{n \to \infty} \frac{f(x)}{1+e^{-nx}} = f(x) \ \ \ \ if \ \ x\in (0,+\infty) $$ Therefore $$\lim_{n \to \infty} \int^{{+ \infty}}_{- \infty} \frac{f(x)e^{nx}}{1+e^{nx}} dx = \int^{+\infty}_{0} f(x) dx$$
I would be thankful if you tell me whether this approach is correct or not. Thanks !
