Exercise in Lebesgue integration.

Question

Find $$ \lim_{n \to \infty} \int_{- \infty}^{\infty} e^{-2|x| \left( 1+ \frac{\arctan(nx)}{\pi} \right)}dx. $$

Thank you.

Answer 1

Note that we have $$e^{-2|x|\left(1+\frac{\arctan(nx)}{\pi}\right)}\le e^{-|x|}$$

and $\int_{-\infty}^\infty e^{-|x|}\,dx<\infty$. Therefore, the dominated convergence theorem guarantees that

$$\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty e^{-2|x|\left(1+\frac{\arctan(nx)}{\pi}\right)}\,dx&=\int_{-\infty}^\infty \lim_{n\to \infty}\left(e^{-2|x|\left(1+\frac{\arctan(nx)}{\pi}\right)}\right)\,dx\\\\ &=\int_{-\infty}^\infty e^{-2|x|\left(1+\frac12\text{sgn}(x)\right)}\,dx\\\\ &=\int_{-\infty}^0 e^{x}\,dx+\int_0^\infty e^{-3x}\,dx\\\\ &=1+\frac13\\\\ &=\frac43 \end{align}$$