Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f \in L^1(-\infty, \infty)$. How to find the limit: $$ \lim_{n \to \infty} \int_{-\infty}^{\infty} \frac{f(x)e^{nx} dx}{1+e^{nx}}. $$ What are the difference between the integration in Riemann integral and Lebesgue integral when we compute integrations? Thank you very much.

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

Hint 1: Apply dominated convergence theorem to compute this integral. Check that all necessary conditions are satisfied.

Hint 2: Prove the following equality $$ \lim\limits_{n\to\infty}\frac{f(x)e^{nx}}{1+e^{nx}}= \begin{cases} f(x)&\quad\text{ if }\quad x>0\\ 0.5 f(0)&\quad\text{ if }\quad x=0\\ 0&\quad\text{ if }\quad x<0\\ \end{cases} $$

share|improve this answer
add comment

For a function $f$ put $f+ = f\vee 0$ and $f^- = (-f)\vee 0$. Then $f = f^+ - f^-$. If $\int f^+$ and $\int f^-$ are finite and $f$ is a.e. continuous, the Riemann and Lebesgue integrals agree. In particular, if $f$ is absolutely integrable in the Riemann sense, it agrees with the Lebesgue integral.

share|improve this answer
Small nit: Riemann integrability requires (local) boundedness of the integrand. –  Harald Hanche-Olsen Aug 1 '12 at 18:35
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.