Ways to evaluate $\lim_{n\to\infty}\int_0^{\frac{1}{n}}\log\left(e^n+e^{-x}\right) \, dx$ as $1$ After I've obtained the indefinite integral 
$$\int\log\left(e^n+e^{-x}\right)\,dx,$$
with Wolfram Alpha, see this code if you need it 
int log(e^n+e^(-x))dx,
and do the calculations at the limits of integration $\int_0^{1/n}$, I get the following limits $$\lim_{n\to\infty}\frac{1}{n}\log\left(\frac{e^n+e^{-\frac{1}{n}}}{e^{n+\frac{1}{n}}+1}\right)+\frac{1}{2n^2}=0$$ and 
$$\lim_{n\to\infty}\operatorname{Li}_2\left(-e^n\right)-\operatorname{Li}_2 \left(-e^{n+\frac{1}{n}}\right)=1.$$ 
I believe that there are no mistakes, thus it implies that $$\lim_{n\to\infty}\int_0^{\frac{1}{n}}\log\left(e^n+e^{-x}\right) \, dx=1.$$
And my question was 

Question. Can you to evaluate (and justify it)  $$\lim_{n\to\infty}\int_0^{\frac{1}{n}}\log\left(e^n+e^{-x}\right) \, dx$$ from a different way? Thanks in advance.


I don't know if makes sense Lebesgue or also Riemann integral. My problem is that I don't understand well the evaluation of the limit of the upper limit and this integrand.
 A: Consider
$$
f_n(x)=\log(e^n+e^{-x})-n=\log(1+e^{-x-n})
$$
Then
$$
0\le\int_0^{1/n}f_n(x)\,dx\le\int_0^{1/n}\log 2\,dx=\frac{\log2}{n}
$$
Thus
$$
\lim_{n\to\infty}\left(\int_0^{1/n}\log(e^n+e^{-x})\,dx-\int_{0}^{1/n}n\,dx\right)=0
$$
A: First, we can write
$$\log(e^n+e^{-x})=n+\log(1+e^{-(n+x)})$$
Then, we have
$$\int_0^{1/n}\log(e^n+e^{-x})\,dx=1+\int_0^{1/n}\log(1+e^{-(n+x)})\,dx$$
So, the problem boils down to showing that $\int_0^{1/n}\log(1+e^{-(n+x)})\,dx\to 0$ as $n\to \infty$.
From the First Mean-Value Theorem for Definite Integrals, there exists a number $\xi_n\in (0,1/n)$ such that
$$\int_0^{1/n}\log(1+e^{-(n+x)})\,dx=\frac1n \log(1+e^{-(n+\xi_n)})$$
Taking the limit as $n\to \infty$ yields the coveted result.
A: For $n$ large, $e^{-x}$ is very nearly $1$ for the entire domain of the integral, while $e^n$ is very large.  So in the limit, the $e^{-x}$ term can be ignored (this is not quite rigorous but it is easy to make it rigorous).
Then you have $$\int_0^{1/n} \log (e^n) \,dx = \int_0^{1/n} n  \,dx = 1$$
So the answer is not at all mysterious.
A: Another approach is to use bounded/dominated convergence: 
$$\int^{1/n}_0  \ln (e^n + e^{-x}) \,dx = \int_0^1 \frac{ \ln (e^n + e^{-y/n})}{n} \, dy\to \int_0^1 1 \, dy.$$   
The limit on the right is because the  nonnegative functions $\ln (e^n + e^{-y/n})/n $  converge pointwise to $1$ as $n\to\infty$ and are  bounded above by  $\ln (2e^n)/n \le \ln 2+1$.   
* NOTE * The pointwise convergence is actually uniform, so we can totally avoid bounded convergence and get the limit from theory of Riemman integration. 
