How to simplify RMM limit of integral? This was posted a while back as an analysis problem from Romanian Mathematical Magazine. It is prove that $\lim _{n\to \infty }\frac{4^n}{\sqrt{n}}\int _0^1\frac{\sqrt{x^2+\frac{1}{e}}}{\left(x^{\frac{1}{n}}+1\right)^{2n}}dx=\sqrt{\frac{\pi }{e}}$. I'm thinking Stirling's approximation and Beta or Hypergeometric functions will come into play but just at this starting point, I don't know how to deal with the different powers of $x$ in the integrand as well as that added $\frac{1}{e}$; the substitutions I've tried haven't given me anything recognizable in terms of special functions or further simplification. Along with this, the indefinite integral itself is not expressible in terms of any known functions which is daunting. I would greatly appreciate any help to simplify this.
 A: Heuristic consideration.
Let's denote $\displaystyle I(n)=\frac{4^n}{\sqrt n}\int _0^1\frac{\sqrt{x^2+\frac{1}{e}}}{\left(x^{\frac{1}{n}}+1\right)^{2n}}dx$.
Making the substitution $x=t^n$
$$I(n)=4^n\sqrt n\int _0^1e^{(n-1)\ln t-2n\ln(1+t)}\sqrt{t^{2n}+\frac{1}{e}}\,dt$$
Making the substitution $x=1-t$
$$=4^n\sqrt n\int _0^1e^{-n\ln \frac{(2-x)^2}{1-x}}\sqrt{(1-x)^{2n}+\frac{1}{e}}\,dx\tag{1}$$
Now we can use Laplace' method. Denoting $f(x)=\ln \frac{(2-x)^2}{1-x}$, we find
$$f'(x)=\frac{x}{(2-x)(1-x)}; f'(x)=0 \,\text{at} \,x=0;\, f''(0)=\frac{1}{2};\, f(0)=\ln 4\tag{2}$$
The main contribution to the integral comes from the region near $x=0$. To find the main asymptotic term, putting (2) into (1),
$$I\sim 4^n\sqrt n\int_0^1e^{-n\ln4-\frac{nx^2}{4}}\sqrt{(1-x)^{2n}+\frac{1}{e}}\,dx$$
$$=\int_0^\sqrt n e^{-\frac{t^2}{4}}\sqrt{(1-t/\sqrt n)^{2n}+\frac{1}{e}}\,dt$$
$(1-t/\sqrt n)^{2n}$ has a non-vanishing contribution only in the narrow interval $0<t\sim\frac{1}{\sqrt n}$; for this interval
$$\int_0^{1/\sqrt n}e^{-\frac{t^2}{4}}\sqrt{(1-t/\sqrt n)^{2n}+\frac{1}{e}}\,dt<\sqrt{1+1/e}\int_0^{1/\sqrt n}dt\sim\frac{const}{\sqrt n}\to0\,\text{at}\, n\to\infty$$
It means that
$$I(n)\sim \frac{1}{\sqrt e}\int_0^\sqrt n e^{-\frac{t^2}{4}}dt\sim\frac{1}{\sqrt e }\int_0^\infty e^{-\frac{t^2}{4}}dt=\sqrt\frac{\pi}{e}$$
