Sufficient bound to conclude limit has certain value. $\lim {\left( {\int_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n}=\frac 1 2 $ I am trying to show that
$$\lim {\left( {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n}=\frac 1 2 $$
Now, this can be done as follows. Using $x\mapsto x^{-1}$ we get that
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  =\int\limits_1^\infty {{{nx^{n-1}} \over {1 + {x^n}}}} \frac{dx}{nx}$$
Upon integrating by parts we get
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  =  - {{\log 2} \over n} + \int\limits_1^\infty  {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} $$
or 
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + \int\limits_1^\infty  {{{\log \left( {1 + {x^n}} \right) - n} \over {n{x^2}}}dx} $$ since $\int_1^\infty x^{-2}dx=1$.
Using $x\mapsto x^{-1}$ once again on the RHS, we get
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + \int\limits_0^1 {{{\log \left( {1 + {x^n}} \right) - n\log x - n} \over n}dx} $$
But since $\int_0^1 \log x=-1=\int_0^1 dx$
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + {1 \over n}\int\limits_0^1 {\log \left( {1 + {x^n}} \right)dx} $$
Now, since for $x\geq 0$, $\log(1+x)\leq x$, we get $$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  \le 1 - {{\log 2} \over n} + {1 \over n}\int\limits_0^1 {{x^n}dx}  = 1 - {{\log 2} \over n} + {1 \over {n\left( {n + 1} \right)}}$$ which means 
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + O\left( {{1 \over {{n^2}}}} \right)$$
Is this enough to conclude that 
$${\left( {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n} = {e^{ - \log 2}} = {1 \over 2}\text{ ? }$$
If so, how? 

Your question can be recast as: if
$$
a_n=1+\frac{a}{n}+O\Bigl(\frac{1}{n^2}\Bigr),
$$
is it true that
$$
\lim_{n\to\infty}a_n^n=e^a?
$$
The answer is yes. Since $\log(1+x)=x+O(x^2)$ as $x\to0$, we have
$$
\log a_n=\frac{a}{n}+O\Bigl(\frac{1}{n^2}\Bigr).
$$
Then
$$
\lim_{n\to\infty}a_n^n=\lim_{n\to\infty}e^{n\log a_n}=e^a.
$$
